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How to replace space in records

Former Member
0 Kudos

Hi All,

My records are in one variable with charecter 200, i want to replace space by %20. How do i do this....

Reg,

Suren

1 ACCEPTED SOLUTION

valter_oliveira
Active Contributor
0 Kudos

DATA: pref TYPE char3 VALUE '%20'.
DATA: aux TYPE char1 VALUE '#'.
TRANSLATE <your_var> USING ' #'.
REPLACE ALL OCCURRENCES OF aux IN <your_var> WITH pref.

Regards,

Valter Oliveira.

8 REPLIES 8

valter_oliveira
Active Contributor
0 Kudos

DATA: pref TYPE char3 VALUE '%20'.
DATA: aux TYPE char1 VALUE '#'.
TRANSLATE <your_var> USING ' #'.
REPLACE ALL OCCURRENCES OF aux IN <your_var> WITH pref.

Regards,

Valter Oliveira.

Former Member
0 Kudos

hi,

use replace for all occurences of that varibale with %20.

syntax is like this.

Replace for all occurances of space in variable with '%20'.

0 Kudos

Hi Subas,

Syntax is showing an error, will appreciate if you can provide me the right Syntax.

Rg,

Suren

Edited by: Suren on Sep 22, 2008 7:44 AM

0 Kudos

Hi Suren,

What is the error u r getting?

Valter's solution given above is working fine.

Regards,

Surinder

0 Kudos

Hi,

for all who seached the solution of the problem in this diskussion like me:

The solution is in first answer of Valter Oliveira.

First translate spaces to sharps and then replace sharps by '%20'.

Direct replace of spaces is not possible, cause only spaces are recognized as empty string, and the result is an infinite loop.

You have only a problem, if you use sharps in your source string ;-(

Regards

Christoph

Former Member
0 Kudos

Suren...

Can you come again with ur ques...Just do u want to eliminate spaces...

Exam "sldjfl slkdjsl sakljkldj slkdjls sdljkkl sdljlj ",

result must be "sdkljdflsjaddjlsjjldjlsjljdlsjldjs"

What is ur exact requirement

Thanks

0 Kudos

Hi, Santosh

Exam "sldjfl slkdjsl sakljkldj slkdjls sdljkkl sdljlj ",

Result Wanted : sldjfl%20slkdjsl%20sakljkldj%20slkdjls%20sdljkkl%20sdljlj ",

Ashoak Narayandas Advani

former_member194797
Active Contributor
0 Kudos

Hi,

I suppose you have not to translate the trailing spaces.

(if you translate the trailing spaces, you have a risk to truncate %20 as %2 or %).

So I should propose this:


data: text(30) value 'this is a test                '.
data: wwlen type i.

write: / text.
wwlen = strlen( text ).
replace all occurrences of ` ` 
in SECTION OFFSET 0 LENGTH wwlen OF text with '%20'.
* CAUTION: use ` ` and not ' ' 
write: / text.