Replace complete words by another string

[Former Member]

Hi,

I'm searching for a possibility to replace complete words inside a string.

For example I want to replace "quick" by "slow":

"The <b>quick</b> brown fox is running quicker then the slow brown fox".

The result shall be:

"The <b>slow</b> brown fox is running quicker then the slow brown fox".

The problem is, that the built in "Replace"-Method would also replace "quicker" by "browner".

Furthermore, words beetween different signs (e.g. , . ; : - ! " ? ') shall also be found as complete word.

I already searched for a FM but didn't find anything. Regular expressesions are also not possible because we are running on 6.40.

Maybe anybody can help me with a simple FM or something else.

Best regards,

Stefan

[Former Member]

Hi,

Try REPLACE ALL OCCURANCES OF statement.

Ashvender

[Former Member]

Hi Ashvender,

as far as I know this would also replace all substrings, or not!?

Stefan

[seshatalpasai_madala]

HI,

Can try with " quick ".

Regards,

Sesh

[Former Member]

Hi Sesh,

that would work. But if there does occur another sign (such as quotes) it wouldn't work...

Stefan

[seshatalpasai_madala]

Hi,

Then may be you can to execute multiple REPLACE statements.

I know its crude but I think no option.

Also may be you can split your string at SPACE and then

execute REPLACE in table.

Regards,

Sesh

[Former Member]

Hi,

Give the line in one string

IF string CN 'quick'.

REPLACE 'quick' WITH 'sloa' INTO field.

ENDIF.

Regards,

Nandha

Reward if it helps

[Former Member]

Hi Stefan,

There is a crude way of doing this....with lots of loops..

You can use the SPLIT statement

SPLIT dobj AT sep INTO

{ {result1 result2 ...} | {TABLE result_tab} }

[IN {CHARACTER|BYTE} MODE].

and break the sentence into words using space as delimiter. Once you have all the words in a internal table, you can loop through it and compare each word with the word you want.

This will solve the problem, but will not be performant enough.

Regards,

Vinodh

[former_member772174]

Hi,

your not replacing 'quick' your replacing ' quick '.

see this:

*----

---

REPORT ZEND_TEST14 MESSAGE-ID zrsp.

data: l_string type string value 'The quick brown fox is running quicker then the slow brown fox'..

replace ' quick ' in l_string with ' slow '.

write l_string.

*----

---

Steffen

[Former Member]

Use

REPLACE <'FIND TEXT'> IN <COMPLETE SEMNTENCE> WITH <'REPLACED TEXT'>.

or

replace all occurances of '.' with ''.

or

REPLACE '.' WITH ' ' INTO in_par-value.

or

var1 = '4444/33'.

REPLACE '/' with SPACE in var1.

Condense var1 no-gaps.

Rewards if useful.

[Former Member]

Hi everybody,

I have played around with different possibilities but nothing seems to be really good. Insertion of space before and after the search-term is not possible because of many other signs which can precede or follow.

I think the only one solution at this time would be to try the different combinations of the "special signs" in combination with the word - but this could slow down the whole application.

I decided not to offer the possibility to replace whole words. Only simple replacements with "REPLACE" will be possible.

Best regards,

Stefan