Technical Interview PHP tutorial

Here is some PHP Interview Questions, that will helpfull to PHP Interview Candidates

1. What is CAPTCHA?

CAPTCHA stands for Completely Automated Public Turing Test to tell Computers and Humans Apart. To prevent spammers from using bots to automatically fill out forms, CAPTCHA programmers will generate an image containing distorted images of a string of numbers and letters. Computers cannot determine what the numbers and letters are from the image but humans have great pattern recognition abilities and will be able to fairly accurately determine the string of numbers and letters. By entering the numbers and letters from the image in the validation field, the application can be fairly assured that there is a human client using it. To read more look here:
http://en.wikipedia.org/wiki/Captcha


2. What is difference between require_once(), require(), include().
Becouse above three function usely use to call a file in another file?

Difference between require() and require_once(): require() includes and evaluates a specific file, while require_once() does that only if it has not been included before (on the same page). So, require_once() is recommended to use when you want to include a file where you have a lot of functions for example. This way you make sure you don't include the file more times and you will not get the "function re-declared" error. Difference between require() and include() is that require() produces a FATAL ERROR if the file you want to include is not found, while include() only produces a WARNING. There is also include_once() which is the same as include(), but the difference between them is the same as the difference between require() and require_once().


3. If you have to work with dates in the following format: "Tuesday, February 14, 2006 @ 10:39 am", how can you convert them to another format, that is easier to use?

The strtotime function can convert a string to a timestamp. A timestamp can be converted to date format. So it is best to store the dates as timestamp in the database, and just output them in the format you like.

So let's say we have
$date = "Tuesday, February 14, 2006 @ 10:39 am";
In order to convert that to a timestamp, we need to get rid of the "@" sign, and we can use the remaining string as a parameter for the strtotime function.

So we have
$date = str_replace("@ ","",$date);
$date = strtotime($date);

now $date is a timestamp
and we can say:

echo date("d M Y",$date);


4. How we know browser properties?

get_browser() attempts to determine the capabilities of the user's browser. This is done by looking up the browser's information in the browscap.ini file.

echo $_SERVER['HTTP_USER_AGENT'] . "

---

\n";

$browser = get_browser();

foreach ($browser as $name => $value) {
echo "$name $value
\n";
}


5. How i will check that user is, logged in or not. i want to make it a function and i want to use in each page and after login i want to go in current page(same page. where i was working)?

For this we can use the session objec($_SESSION)t. When the user login with his/ her user name and password, usually we check those to ensure for correctness. If that user name and password are valid one then we can store that user name in a session and then we can very that session variable has been set or not in a single files and we can include that file in all pages.

6. How i can get ip address?

We can use SERVER var $_SERVER['SERVER_ADDR'] and getenv("REMOTE_ADDR") functions to get the IP address.

7. What is differenc between mysql_connect and mysql_pconnec?

mysql_pconnect establishes a persistent connection. If you don't need one (such as a website that is mostly HTML files or PHP files that don't call the db) then you don't need to use it. mysql_connect establishes a connection for the duration of the script that access the db. Once the script has finished executing it closes the connection. The only time you need to close the connection manually is if you jump out of the script for any reason.

If you do use mysql_pconnect. You only need to call it once for the session. That's the beauty of it. It will hold open a connection to the db that you can use over and over again simply by calling the resource ID whenever you need to interact with the db.

8. What is the difference between echo and print statement?

There is a slight difference between print and echo which would depend on how you want to use the outcome. Using the print method can return a true/false value. This may be helpful during a script execution of somesort. Echo does not return a value, but has been considered as a faster executed command. All this can get into a rather complicated discussion, so for now, you can just use whichever one you prefer.

9. How to make a download page in own site, which i can know that how many file has been loaded by particular user or particular ipaddress?

We can use hyperlink having URL where file are kept. and we only allow regisetered user to download. from session of user we can get the user detail

Read more »

Technical Interview "C" Q & A Part 2

Predict the output or error(s) for the following:

1. main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i
{
printf(“%s\n”,x);
x++;
}
}

Answer:
(blank space)
irl
rl
l

Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The strlen function returns
the length of the string, thus n has a value 4. The next statement assigns value at the nth location
(‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the
string after each iteration it increments it starting position. Loop starts from 0 to 4. The first
time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints
from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop
terminates.

2. int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}

Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), ,

Explanation:
asserts are used during debugging to make sure that certain conditions are satisfied. If assertion
fails, the program will terminate reporting the same. After debugging use,
#undef NDEBUG
and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.

3. main()
{
int i=-1;
+i;
printf("i = %d, +i = %d \n",i,+i);
}

Answer:
i = -1, +i = -1

Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has
no effect in the expressions (hence the name dummy operator).

4. What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).

5. what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);

Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.

6. main()
{
char name[10],s[12];
scanf(" \"%[^\"]\"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches with a quotation mark and then
it reads all character upto another quotation mark.

7. What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for != NULL.

8. main()
{
main();
}

Answer:
Runtime error : Stack overflow.

Explanation:
main function calls itself again and again. Each time the function is called its return address is stored in
the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime.
So it terminates the program and results in an error.

9. main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}

Answer:
Compiler error (at line number 4): size of v is Unknown.

Explanation:
You can create a variable of type void * but not of type void, since void is an empty type. In the second
line you are creating variable vptr of type void * and v of type void hence an error.

10. main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}

Answer:
2 5 5

Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof
the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character).
The third sizeof is similar to the second one.

11. main()
{
char not;
not=!2;
printf("%d",not);
}

Answer:
0

Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value
is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

12. #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}

Answer:
TRUE

Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean
value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

13. main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}

Answer:
1==1 is TRUE

Explanation:
When two strings are placed together (or separated by white-space) they are concatenated (this is called as
"stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: )
evaluates to "TRUE".

14. main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}

Answer:
2000 is a leap year

Explanation:
An ordinary program to check if leap year or not.

15. #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}

Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})

Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of
the type arr2. But it is not the case of arr1. Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for declaring new types.

16. int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}

Answer:
30,20,10

Explanation:
'{' introduces new block and thus new scope. In the innermost block i is declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20
and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it.
After compilation is over the linker resolves it to global variable i (since it is the only variable visible there).
So it prints i's value as 10.

17. main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}

Answer:
10

Explanation:
The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is
lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space,
*j prints the value stored in i since j points i.

18. main()
{
int i=-1;
-i;
printf("i = %d, -i = %d \n",i,-i);
}

Answer:
i = -1, -i = 1

Explanation:
-i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i.
After that the value of the expression -i = -(-1) is printed.

19. #include
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}

Answer:
Compiler error

Explanation:
i is a constant. you cannot change the value of constant

20. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}

Answer:
garbagevalue..1

Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to access the third 2D
(which you are not declared)
it will print garbage values. *q=***a starting address of a is assigned integer pointer.
now q is pointing to starting address of a.if you print *q meAnswer:it will print first
element of 3D array.

21. #include
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}

Answer:
hello 5

Explanation:
if you declare i as register compiler will treat it as ordinary integer and it will take
integer value. i value may be stored either in register or in memory.

22. main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}

Answer:
11

Explanation:
the expression i+++j is treated as (i++ + j)

Read more »

Technical interview "C " Q & A

Predict the output or error(s) for the following:


1. void main()
{
int const * p=5;
printf("%d",++(*p));
}

Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}

Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea. Generally array name is
the base address for that array. Here s is the base address. i is the index number/displacement from the
base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C
it is same as s[i].

3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}

Answer:
I hate U

Explanation:
For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending
on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long
double takes 10 bytes. So float stores 0.9 with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) .

4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}

Answer:
5 4 3 2 1

Explanation:
When static storage class is given, it is initialized once. The change in the value of a static variable is
retained even between the function calls. Main is also treated like any other ordinary function, which can
be called recursively.

5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
}

Answer:
2 2 2 2 2 2 3 4 6 5

Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c ,
the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6. main()
{
extern int i;
i=20;
printf("%d",i);
}

Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and that address will be
given to the current program at the time of linking. But linker finds that no other variable of name i is
available in any other program with memory space allocated for it. Hence a linker error has occurred .

7. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}

Answer:
0 0 1 3 1

Explanation :
Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority
over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this
expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator
always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of
other variables are also incremented by 1.

8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
}

Answer:
1 2

Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one
byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store
the address of the character pointer sizeof(p) gives 2.

9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}

Answer :
three

Explanation :
The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10. main()
{
printf("%x",-1<<4);
}

Answer:
fff0

Explanation :
-1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled
with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

11. main()
{
char string[]="Hello World";
display(string);
}
void display(char *string)
{
printf("%s",string);
}

Answer:
Compiler Error : Type mismatch in redeclaration of function display

Explanation :
In third line, when the function display is encountered, the compiler doesn't know anything about the function
display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the
actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile
time error occurs.

12. main()
{
int c=- -2;
printf("c=%d",c);
}

Answer:
c=2;

Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator
(eg., i--). 2 is a constant and not a variable.

13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}

Answer:
sizeof(i)=1

Explanation:
Since the #define replaces the string int by the macro char

14. main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}

Answer:
i=0

Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol. ! is a unary logical operator.
!i (!10) is 0 (not of true is false). 0>14 is false (zero).

#include
main()
{
char s[]={'a','b','c','\n','c','\0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}

Answer:
77

Explanation:
p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is
incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11.
++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).

16. #include
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}

Answer:
SomeGarbageValue---1

Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not
declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q
is pointing to starting address of a. If you print *q, it will print first element of 3D array.

17. #include
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}

Answer:
Compiler Error

Explanation:
You should not initialize variables in declaration

18. #include
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}

Answer:
Compiler Error

Explanation:
The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through
the instance of structure xx, which needs an instance of yy to be known. If the instance is created after
defining the structure the compiler will not know about the instance relative to xx. Hence for nested
structure yy you have to declare member.

19. main()
{
printf("\nab");
printf("\bsi");
printf("\rha");
}

Answer:
hai

Explanation:
\n - newline
\b - backspace
\r - linefeed

20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}

Answer:
45545

Explanation:
The arguments in a function call are pushed into the stack from left to right. The evaluation is by
popping out from the stack. and the evaluation is from right to left, hence the result.

21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}

Answer:
64

Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and *
has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='\0') ++*p++;
printf("%s %s",p,p1);
}

Answer:
ibj!gsjfoet

Explanation:
++*p++ will be parse in the given order
Ø *p that is value at the location currently pointed by p will be taken
Ø ++*p the retrieved value will be incremented
Ø when ; is encountered the location will be incremented that is p++ will be executed
Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p
and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is
converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points
to p thus p1doesnot print anything.

23. #include
#define a 10
main()
{
#define a 50
printf("%d",a);
}

Answer:
50

Explanation:
The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24. #define clrscr() 100
main()
{
clrscr();
printf("%d\n",clrscr());
}

Answer:
100

Explanation:
Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement
of clrscr() to 100 occurs.The input program to compiler looks like this :
main()
{
100;
printf("%d\n",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any problem

25. main()
{
printf("%p",main);
}

Answer:
Some address will be printed.

Explanation:
Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf specifies that
the argument is an address. They are printed as hexadecimal numbers.

Read more »

Tough interview questions on EJB

1. How EJB Invocation happens? - Retrieve Home Object reference from Naming Service via JNDI. Return Home Object reference to the client. Create me a new EJB Object through Home Object interface. Create EJB Object from the Ejb Object. Return EJB Object reference to the client. Invoke business method using EJB Object reference. Delegate request to Bean (Enterprise Bean).
2. Is it possible to share an HttpSession between a JSP and EJB? What happens when I change a value in the HttpSession from inside an EJB? - You can pass the HttpSession as parameter to an EJB method, only if all objects in session are serializable.This has to be consider as passed-by-value, that means that it’s read-only in the EJB. If anything is altered from inside the EJB, it won’t be reflected back to the HttpSession of the Servlet Container.The pass-by-reference can be used between EJBs Remote Interfaces, as they are remote references. While it is possible to pass an HttpSession as a parameter to an EJB object, it is considered to be bad practice in terms of object-oriented design. This is because you are creating an unnecessary coupling between back-end objects (EJBs) and front-end objects (HttpSession). Create a higher-level of abstraction for your EJBs API. Rather than passing the whole, fat, HttpSession (which carries with it a bunch of http semantics), create a class that acts as a value object (or structure) that holds all the data you need to pass back and forth between front-end/back-end. Consider the case where your EJB needs to support a non HTTP-based client. This higher level of abstraction will be flexible enough to support it.
3. The EJB container implements the EJBHome and EJBObject classes. For every request from a unique client, does the container create a separate instance of the generated EJBHome and EJBObject classes? - The EJB container maintains an instance pool. The container uses these instances for the EJB Home reference irrespective of the client request. while refering the EJB Object classes the container creates a separate instance for each client request. The instance pool maintenance is up to the implementation of the container. If the container provides one, it is available otherwise it is not mandatory for the provider to implement it. Having said that, yes most of the container providers implement the pooling functionality to increase the performance of the application server. The way it is implemented is, again, up to the implementer.
4. Can the primary key in the entity bean be a Java primitive type such as int? - The primary key can’t be a primitive type. Use the primitive wrapper classes, instead. For example, you can use java.lang.Integer as the primary key class, but not int (it has to be a class, not a primitive).
5. Can you control when passivation occurs? - The developer, according to the specification, cannot directly control when passivation occurs. Although for Stateful Session Beans, the container cannot passivate an instance that is inside a transaction. So using transactions can be a a strategy to control passivation. The ejbPassivate() method is called during passivation, so the developer has control over what to do during this exercise and can implement the require optimized logic. Some EJB containers, such as BEA WebLogic, provide the ability to tune the container to minimize passivation calls. Taken from the WebLogic 6.0 DTD -”The passivation-strategy can be either “default” or “transaction”. With the default setting the container will attempt to keep a working set of beans in the cache. With the “transaction” setting, the container will passivate the bean after every transaction (or method call for a non-transactional invocation).
6. What is the advantage of using Entity bean for database operations, over directly using JDBC API to do database operations? When would I use one over the other? - Entity Beans actually represents the data in a database. It is not that Entity Beans replaces JDBC API. There are two types of Entity Beans Container Managed and Bean Mananged. In Container Managed Entity Bean - Whenever the instance of the bean is created the container automatically retrieves the data from the DB/Persistance storage and assigns to the object variables in bean for user to manipulate or use them. For this the developer needs to map the fields in the database to the variables in deployment descriptor files (which varies for each vendor). In the Bean Managed Entity Bean - The developer has to specifically make connection, retrive values, assign them to the objects in the ejbLoad() which will be called by the container when it instatiates a bean object. Similarly in the ejbStore() the container saves the object values back the the persistance storage. ejbLoad and ejbStore are callback methods and can be only invoked by the container. Apart from this, when you use Entity beans you dont need to worry about database transaction handling, database connection pooling etc. which are taken care by the ejb container.
7. What is EJB QL? - EJB QL is a Query Language provided for navigation across a network of enterprise beans and dependent objects defined by means of container managed persistence. EJB QL is introduced in the EJB 2.0 specification. The EJB QL query language defines finder methods for entity beans with container managed persistenceand is portable across containers and persistence managers. EJB QL is used for queries of two types of finder methods: Finder methods that are defined in the home interface of an entity bean and which return entity objects. Select methods, which are not exposed to the client, but which are used by the Bean Provider to select persistent values that are maintained by the Persistence Manager or to select entity objects that are related to the entity bean on which the query is defined.
8. Brief description about local interfaces? - EEJB was originally designed around remote invocation using the Java Remote Method Invocation (RMI) mechanism, and later extended to support to standard CORBA transport for these calls using RMI/IIOP. This design allowed for maximum flexibility in developing applications without consideration for the deployment scenario, and was a strong feature in support of a goal of component reuse in J2EE. Many developers are using EJBs locally, that is, some or all of their EJB calls are between beans in a single container. With this feedback in mind, the EJB 2.0 expert group has created a local interface mechanism. The local interface may be defined for a bean during development, to allow streamlined calls to the bean if a caller is in the same container. This does not involve the overhead involved with RMI like marshalling etc. This facility will thus improve the performance of applications in which co-location is planned. Local interfaces also provide the foundation for container-managed relationships among entity beans with container-managed persistence.
9. What are the special design care that must be taken when you work with local interfaces? - It is important to understand that the calling semantics of local interfaces are different from those of remote interfaces. For example, remote interfaces pass parameters using call-by-value semantics, while local interfaces use call-by-reference. This means that in order to use local interfaces safely, application developers need to carefully consider potential deployment scenarios up front, then decide which interfaces can be local and which remote, and finally, develop the application code with these choices in mind. While EJB 2.0 local interfaces are extremely useful in some situations, the long-term costs of these choices, especially when changing requirements and component reuse are taken into account, need to be factored into the design decision.
10. What happens if remove( ) is never invoked on a session bean? - In case of a stateless session bean it may not matter if we call or not as in both cases nothing is done. The number of beans in cache is managed by the container. In case of stateful session bean, the bean may be kept in cache till either the session times out, in which case the bean is removed or when there is a requirement for memory in which case the data is cached and the bean is sent to free pool.
11. What is the difference between Message Driven Beans and Stateless Session beans? - In several ways, the dynamic creation and allocation of message-driven bean instances mimics the behavior of stateless session EJB instances, which exist only for the duration of a particular method call. However, message-driven beans are different from stateless session EJBs (and other types of EJBs) in several significant ways: Message-driven beans process multiple JMS messages asynchronously, rather than processing a serialized sequence of method calls. Message-driven beans have no home or remote interface, and therefore cannot be directly accessed by internal or external clients. Clients interact with message-driven beans only indirectly, by sending a message to a JMS Queue or Topic. Only the container directly interacts with a message-driven bean by creating bean instances and passing JMS messages to those instances as necessary. The Container maintains the entire lifecycle of a message-driven bean; instances cannot be created or removed as a result of client requests or other API calls.
12. How can I call one EJB from inside of another EJB? - EJBs can be clients of other EJBs. It just works. Use JNDI to locate the Home Interface of the other bean, then acquire an instance reference, and so forth.
13. What is an EJB Context? - EJBContext is an interface that is implemented by the container, and it is also a part of the bean-container contract. Entity beans use a subclass of EJBContext called EntityContext. Session beans use a subclass called SessionContext. These EJBContext objects provide the bean class with information about its container, the client using the bean and the bean itself. They also provide other functions. See the API docs and the spec for more details.

Read more »