detect asterisk from parameter (user input)

[Former Member]

Hi all,

How do you detect asterisks from an inputted parameter of type string?

The user is supposed to use asterisks on the left, right, or both sides of the word/s that they type in the parameter field.

Parameters: p_txt50 type string.

This is what i want to do:
If p_txt50 has asterisk to the right (ex. computer*)
[code goes here]
If p_txt50 has asterisk to the left  (ex. *computer)
[code goes here]
If p_txt50 has asterisk on both sides  (ex. *computer*)
[code goes here]
Thank you very much :)

Edited by: mrpena on Jul 24, 2009 7:14 AM

Edited by: mrpena on Jul 24, 2009 7:15 AM

[former_member209217]

Hi,

U can Do it using String Operations.

Try using CP (Contains Pattern ).

regards,

Lakshman.

[sridhar_meesala]

Hi,

You can achieve it using

CP - Contains Pattern or

NP - Contains No Pattern

Thanks,

Sri.

[Former Member]

Thank you for the replies. I found the CP documentation, but I couldn't find how to use it properly.

Can you elaborate more (on the actual syntax) that I could use?

[Former Member]

if p_text CP '*'. " This means ur parameter has any * then execute ur code

your code.

endif.

Here param is the parameter name in ur program.

Edited by: vijetasap on Jul 24, 2009 7:29 AM

[viquar_iqbal]

Hi

DATA: f1(5) TYPE c VALUE 'aBcde',

      f2(5) TYPE c VALUE '*b*'.

IF f1 CP f2 .
   WRITE: /  'Value found, SY-FDPOS=', sy-fdpos.
ELSE.
   WRITE: /  'Value not found, SY-FDPOS=', sy-fdpos.
ENDIF.

Hope this clears the issue.

Thanks

Viquar Iqbal

[Former Member]

Hi,

Try Like below;

Data : lv_len type i.

lv_len = strlen( p_txt50 ).
lv_len = lv_len - 1.
if p_txt50+0(1) EQ '*' AND p_txt50+lv_len(1) EQ '*'.
"Contains Asterik in both sides.
else if p_txt50+0(1) EQ '*' AND p_txt50+lv_len(1) NE '*'.
"Contains Asterik in front
else if p_txt50+0(1) NE '*' AND p_txt50+lv_len(1) EQ '*'.
"Contains Asterik in last
endif.

Check this and revert back.

Regards

Karthik D

[former_member585060]

Hi,

Do as below

DATA : l TYPE i.

PARAMETERS : p_var50 TYPE string.

AT SELECTION-SCREEN ON p_var50.

  l = STRLEN( p_var50 ).
  l = l - 1.

START-OF-SELECTION.

  IF p_var50+0(1) = '*' AND p_var50+l(1) = '*'.

    WRITE : 'Both ends'.

  ELSE.
    IF p_var50+0(1) = '*'.

      WRITE : 'Starting'.

    ELSE.
      IF p_var50+l(1) = '*'.

        WRITE : 'End'.

      ENDIF.
    ENDIF.

Regards

Bala Krishna

[Former Member]

Also, you will have to use -

---

> strlen(variable) to check the string length..

and str1 = str2+2(5).

What you can do is...

1) to check whether the last char is '*', calculate string length, minus 1 from it(v1), and put in the statement..

str1 = str2+v1(v2) " where v2 = 1.

now check whether it's '*' or not in str1.

2) Use similar logic for

syntaxes for string manipulations..

Exactly as Kartik has suggested!

Regards,

Sumit Nene.

Edited by: Sumit Nene on Jul 24, 2009 7:32 AM

[former_member386202]

Hi,

Use SEARCH Statement. Refer Following code

SEARCH p_text FOR '*'.

IF sy-subrc EQ 0.

ENDIF.

Ragards,

Prashant

[Former Member]

HI,

Declar two variables say var1 and var2. Now check if the first character of the entered string is * and the last character is * then process the code when fisrt and last characters are *, else check if only fist character is * or last character is *.

Following is s pseudo code

var1 = p_txt50(0)+1.
v_len = strlen( p_txt50  ).
v_len = v_len - 1.
var2 = p_txt50(v_len)+1.
if var1 = '' and var2 = '*'.
insert code for *computer*
elseif var1 = '*'.
insert code for *computer
elseif var2 = '*'.
insert code for computer*.
else
insert code for non of the abpve condition.
endif.

Hope this helps.

Regards,

Sachin

[Former Member]

This message was moderated.

[Former Member]

ok thanks for the replies. i'm now trying out the many codes and may take some time. i'll reply right back and tell what happens.

[Former Member]

Hi everyone.

Thank you for responding. My program works great. Thanks more power