Replace % in the end of string with Regex?

user_11637777 0
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I want to remove " %" at the end of some text. I'd like to do it with a regular expression, because ABAP does not easily handle text at the end of a string.

DATA lv_vtext TYPE c LENGTH 10 VALUE 'TEST %'.
REPLACE REGEX ' %$' IN lv_vtext WITH ''.

But it does not replace anything. When I leave out "$" the text will be removed as expected, but I fear it might find more occurrences than wanted.

I experimented with \z or \Z instead of $, but to no avail.

abap regex
2 Answers

This answer is about an alternative way without REGEX. POSIX regular expressions are quite slow, moreover some people are reluctant to use it, so if you're not completely closed to do it in normal ABAP:

lv_vtext = COND #( WHEN contains( val = lv_vtext end = ` %` ) 
           THEN substring( val = lv_vtext len = strlen( lv_vtext ) - 2 )
           ELSE lv_vtext ).

Code with context:

DATA(lv_vtext) = `test %`.
lv_vtext = COND #( WHEN contains( val = lv_vtext end = ` %` ) 
           THEN substring( val = lv_vtext len = strlen( lv_vtext ) - 2 )
           ELSE lv_vtext ).
ASSERT lv_vtext = `test`.
user_91502700

You can use

REPLACE REGEX '\s%\s*$' IN lv_vtext WITH ''

The benefit of using \s is that it matches any Unicode whitespace chars. The \s*$ matches any trailing (white)spaces that you might have missed.

The whole pattern matches

  • \s - any whitespace
  • % - a % char
  • \s* - zero or more whitespaces
  • $ - at the end of string.
user_38329700
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