topic Re: Problem in IF condition in Application Development and Automation Discussions

Problem in IF condition

Former Member — Mon, 06 Jul 2009 12:24:53 GMT

Dear All,

Please see the below code.


tables  : LFM1.
data :  w_menge(16) type c.

LFM1-LFABC = 'X'.
W_MENGE = '0.499'.

IF LFM1-LFABC NE SPACE AND W_MENGE NE 0.

write :'SUCCESS'.

else.

write : 'FAIL'.
endif.

In the above code if the value of W_MENGE >= 0.5, the If condition gets satisfied and SUCCESS is displayed on screen.

But if the value of W_MENGE < 0.5, the codtion fails, and FAIL is displayed on screen.

Can any one please tell me the reason?

Regards

Arindam

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 12:27:33 GMT

Dear,

You are using Data Type C for Your Qualityt field.

How could it do comparisons?

Regds,

Anil

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 12:28:59 GMT

U use the below code.---

if it_table-menge GT 0.5.

write 'SUCCESS'.

else.

write 'FAIL'.

endif.

Re: Problem in IF condition

former_member434229 — Mon, 06 Jul 2009 12:30:28 GMT

Hi,

Check whther you have selected 'Fixed Point Arithmetic' while creating the program.

To check it Open the program - > Goto -> Attributes. Check it is not selected.

Regards,

Ni3

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 12:32:10 GMT

Hi,

Declare W_MENGE as follows.

DATA: W_MENGE type P DECIMALS 3. " NOTE dont use FLOAT because the value will be changed

This will work

Regards,

Smart Varghese

Edited by: smartvarghese on Jul 6, 2009 2:42 PM

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 12:32:12 GMT

Hi,

define w_menge as float ie f.

LFABC = 'x'.

W_MENGE = '0.499'.

IF LFABC NE SPACE AND W_MENGE NE 0.

write :'SUCCESS'.

else.

write : 'FAIL'.

endif.

thanks.

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 12:32:21 GMT

HI,

" Have 0 in quotes then you get the success message..
IF LFM1-LFABC NE SPACE AND W_MENGE NE '0'.

But still it is wrong..instead taking the char take the float type and compare the field. This is the right way.

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 12:43:10 GMT

Thanks Avinash. I did the same put quotes.

But it looks strange why it is working fine with value 0.5 or any value greater than that.

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 12:44:32 GMT

Hi,

You try using 'NOT INITIAL' instead of NE 0 which is shown below

tables  : LFM1.
data :  w_menge(16) type c.
 
LFM1-LFABC = 'X'.
W_MENGE = '0.499'.
 
IF LFM1-LFABC NE SPACE AND W_MENGE IS NOT INITIAL."this works if the value is not blank
    " if you wish to check value non zero, then you can do as said above that is
    "IF LFM1-LFABC NE SPACE AND W_MENGE NE '0'.
 
write :'SUCCESS'.
 
else.
 
write : 'FAIL'.
endif.

Regards,

Manne.

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 13:01:59 GMT

I got the solution for this, BUt still dont understand reason behind it.

Why is it working fine with value 0.5 or any value greater than that?

Can anyone please let me know.

Regards

Arindam

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 13:08:09 GMT

That is because first you were trying to compare a character field with an integer value.

if v1 ne 0.

here 0 is taking as integer and for char variable v1 it is automatically converting to integer value for comparision.

so < 0.5 convert to integer is always 0.

and > 0.5 is convert to integer as 1.

so it is converting '0.499' to 0 and give you the wrong result.

But when you add qoutes ( '0') it became a comparision between two characters field so no conversion was made between character to int.

Now for that reason you are getting the correct result.

Regards

Shiba Prasad Dutta

Re: Problem in IF condition

Former Member — Mon, 06 Jul 2009 13:11:10 GMT

The answer is easy: rounding.

With values smaller than 0.5 it rounds to 0. Values greater than 0.5 round to 1.