Search and replace last 4 caracteres in a string

[Former Member]

Hi Guys,

First and foremost moderators, I do know abap, have been working for many years with it.

Had an issue where I lost partially my memory and only now after a couple of years I am back in the workforce.

So please do not delete my post. Would be very helpful to get some answers.

So, in one of the three interfaces I am working on I need to:

Search for 'av.' in a string.

If 'av.' is found, it needs to be replaced by 'avenue'

If nothing is found string stays as it is.

I have a long code and for some reason my mind only remembers the difficult stuff ...

On my code:

Data: V_string TYPE String.

And after on the code:

V_string = ls_itab-street.

Thank you all!

F

[Former Member]

Hi Guys,

First and foremost moderators, I do know abap, have been working for many years with it.

Had an issue where I lost partially my memory and only now after a couple of years I am back in the workforce.

So please do not delete my post. Would be very helpful to get some answers.

So, in one of the three interfaces I am working on I need to:

Search for 'av.' in a string.

If 'av.' is found, it needs to be replaced by 'avenue'

If nothing is found string stays as it is.

I have a long code and for some reason my mind only remembers the difficult stuff ...

On my code:

Data: V_string TYPE String.

And after on the code:

V_string = ls_itab-street.

Thank you all!

F

[Former Member]

DATA: T(10) VALUE 'av.defghij',
      STRING LIKE T,
      STR1(3) VALUE 'av.',
      STR2(6) VALUE 'avenue',

STRING = T.

IF STRING CS STR1

REPLACE STR1 WITH STR2 INTO STRING.

ENDIF.

[anupam_anand]

Hi Santhoshini,

There may be some issues here. The requirement here is to search and replace in the last 4 characters. Also, the direct 'REPLACE' may eliminate the 'av.' from other portions of the text also if present which is not correct.

Thanks,

Anupam

[anupam_anand]

Hi Fafa,

Please find the code below:

DATA: lv_str1 TYPE c LENGTH 40,

      lv_str2 TYPE c LENGTH 40,

      lv_line TYPE i.

lv_str1 = 'Anupam Anand av.'.

lv_line = strlen( lv_str1 ).

lv_line = lv_line - 4.

lv_str2 = lv_str1+lv_line(4).

IF lv_str2 ca 'av.'.

REPLACE ALL OCCURRENCES OF 'av.' in lv_str2 WITH 'avenue'.

lv_str1+lv_line(4) = ''.

CONCATENATE lv_str1 lv_str2 INTO lv_str1.

ENDIF.

WRITE: lv_str1.

Hope this helps

Thanks,

Anupam

[Former Member]

Hi Fafa,

there you go. The easiest solution imho is using regular expressions. A very useful thing to learn.

REPORT  zav_replace_test.

DATA: lt_testdata TYPE TABLE OF string,

      lv_string LIKE LINE OF lt_testdata.

lv_string = 'Text with av. in the middle'.

INSERT lv_string INTO TABLE lt_testdata.

lv_string = 'Navigation abbreviated as Nav.'.

INSERT lv_string INTO TABLE lt_testdata.

lv_string = 'Navigation abbreviated as Nav. in a sentance'.

INSERT lv_string INTO TABLE lt_testdata.

lv_string = 'Palm av.'.

INSERT lv_string INTO TABLE lt_testdata.

LOOP AT lt_testdata INTO lv_string.

  WRITE: / 'Before replace:', lv_string.

  " Explanation of the regex

  " \s one white space

  " av the text av

  " \. a literale .

  " $ the end of the string

  REPLACE REGEX '\sav\.$' IN lv_string WITH ' Avenue'.

  WRITE: / 'After replace:', lv_string.

ENDLOOP.

[Former Member]

Find your words whatever you want to search by using 'CS'. Take that word into a variable and replace it by using replace command.

Regards

Vivek