Identifying User Exits

[Former Member]

How can we identify the user exist?

[Former Member]

hai Sandeep Tiwari,

I am Genaralising This For U.

It Is ver EASY Procees.

Just Go To SE84

Give The PRogram Name Or PAckage name In Which VL02N is present.

Then Click On Display.

Then Click On Enhancements.

Then U will Get USer Exits Aswell As BADI 's

IF Found helpful Reward.

Regards

Eshwar

[SantoshKallem]

we have several methods.

find the package name of the transaction. hope u know how to.

run t-code, system--status. double click on program, check the attribute of that program.

then goto t-code smod or cmod, extras--> find.

give the package name and execute.

other method.

se81 application hierarchy.

select the hierarcy upto u r knowledge where the exit u required.

press the button set focus on tool bar. it changes to yellow color.

select the push button(components, not sure..) on tool bar.

expant the enhancements or badis which ever u want, double click on it and execute..

one more..

open the program and fin for CALL CUSTOMER.

one more.

open the spro.

press ctrl+f and search for enhancements.

regards.

santhosh reddy

[Former Member]

You can find the exits in many ways.

Get the developement class of the transaction code from SE93.

Then go to transaction SMOD and press F4 and give the Development class in Press ENTER, then it will show all the exits available for that transaction.

Or

Go to transaction SE80 and give the development class name and then expand the enhancements tree view then you will get the list of userexits for that corresponding dev class.

To check for the suitable badi, Go to transaction SE24 and give the Object type as CL_EXITHANDLER and click on Display button.

Double click on method GET_INSTANCE, put a break point on case statement.

Execute your transaction and look for variable EXIT_NAME at your scenario then It will show the available badi's.

Thanks,

Srinivas

[Former Member]

Use the transaction COMD and for that respective development class exits are available with + sign.