contains #

[Former Member]

Hi,

I am trying to write the following code

if l_betrg ca '#'.

g_error = 'X'.

endif.

but even if l_betrg contains #, it is not going to g_error = 'X'

Could you please tell me where i am going wrong..I need to identify if l_betrg contains #. l_betrg is defined as

data: l_betrg(13) type c.

Thanks

Pranati

[Former Member]

try this..

if l_betrg ca <b>CL_ABAP_CHAR_UTILITIES=>HORIZONTAL_TAB.</b>

g_error = 'X'.

endif.

<b>OR</b>

if l_betrg ca <b>CL_ABAP_CHAR_UTILITIES=>CR_LF.</b>

g_error = 'X'.

endif.

[Former Member]

try this..

if l_betrg ca <b>CL_ABAP_CHAR_UTILITIES=>HORIZONTAL_TAB.</b>

g_error = 'X'.

endif.

<b>OR</b>

if l_betrg ca <b>CL_ABAP_CHAR_UTILITIES=>CR_LF.</b>

g_error = 'X'.

endif.

[Former Member]

Hi chandrashekar,

Just curious.. y did not my code work?

[Former Member]

it depends on the unicode , even i am not sure about that...

it works sometimes and sometimes not...

[amit_khare]

Try this -

str1 type string value '#.

str2 type string.

move l_betrg to str2

if str2 cs str1.

g_error = 'X'.

endif.

Regards,

Amit

reward all helpful replies.

[Former Member]

Your test case scenario doesn't happen... There might be a mistake in the content of the Variable... Please check the code.

Regards,

Pavan P.

[Former Member]

Hi Praniti,

Use CP instead of CA.

Reward if useful!

[seshatalpasai_madala]

Hi,

Check in debug mode if this actually contains '#' may its a case that it contains some value which is getting displayed as #.

Regards,

Sesh

[former_member188827]

i tried ur code.its working absolutely fine!!

data i_betrg(13) type c.

data g_error(1).

i_betrg = '#1234'.

if i_betrg ca '#'.

g_error = 'X'.

endif.

write g_error.

rgds

[Former Member]

Hi,

I think CA operator checks to see whether any letter in field1 is in field2.

Reward if helpful

Regards

Raghavendra.D.S

[Former Member]

Hi,

actual probem is "#" is not a value.... its just showing the deliminator..... u cant use in the condition.... actually sap use it when its unidentified value

1. sign just determine the space or deliminator.....its actuall value is not a "#"... thats why your condition is being false..

regards,

shardul shah

Message was edited by:

Shardul Shah