Variable Exit Code i_step = 2

Question

Hi There
I have an issue with the code of type i_step = 2. What i want to do is to check a table and if a certain value is in that table i want to include a restriction for all values for the caracteristic on which the variable is based.
My code looks like this:
IF zuser_auth-comp_code = '*'.
          ls_range-low = '*'.
          ls_range-sign = 'I'.
          ls_range-opt = 'CP'.
          APPEND ls_range to e_t_range.
          EXIT.
ENDIF.
THE CP = * restriction is used without problems for i_step = 0 variables but it does not work for i_step = 2. Does anybody know why or know another way to restrict to all values in a i_step = 2 variable exit??
I hope that you can help me.
/Chris

Answer 1

Hi again
You are right the I_step = 2 is executed after execution of the query. When i debug the exit my exit variable code is executed but the result is that the program ends with an illegal state.Below i have inserted the complete code.
WHEN 'ZCECCST2'.  "Authorization check for Company Code
    IF i_step = '2'.
      READ TABLE i_t_var_range into loc_var_range with key IOBJNM = '0COMP_CODE'.
      if sy-subrc <> '0'.
      SELECT * FROM ZUSER_AUTH WHERE
          username = l_uname.
      IF zuser_auth-comp_code = '*'.
          SELECT comp_code from /bi0/mcomp_code into ls_range-low.
          ls_range-low = '*'.
          ls_range-sign = 'I'.
          ls_range-opt = 'CP'.
          APPEND ls_range to e_t_range.
          ENDSELECT.
          EXIT.
      ELSE.
          ls_range-low = zuser_auth-comp_code.
          ls_range-sign = 'I'.
          ls_range-opt = 'EQ'.
          if ls_range-low <> ''.
            APPEND ls_range to e_t_range.
          endif.
      ENDIF.
      ENDSELECT.
      ELSE.
          ls_range-low = loc_var_range-low.
          ls_range-sign = 'I'.
          ls_range-opt = 'EQ'.
          APPEND ls_range to e_t_range.
      ENDIF.
  endif.
/Chris