replace a substring by position

[atamo]

I want to replace a string second character with somthing else, e.g. :

set cmd /E:ON
set var=asdfg
echo !var:!var:~1,1!=OK!

unfortunately don't replace s with OK , the code display

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!var:s=OK!

but I want

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aOKdfg

thank's your help .

[Squashman]

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H:\>set var=asdfg

H:\>set var=%var:s=OK%

H:\>echo %var%
aOKdfg

[dbenham]

Squashman's solution searches for a substring and replaces it with something else.

You want to replace based on position, not a substring.

You attempted to use delayed expansion (which is a good idea if you are dealing with special characters like ^ & > | < etc.) But you did not enable delayed expansion properly. Your substring syntax was also incorrect.

You can't really replace a character by position like that. Instead you have to build a new string by concatenating the beginning substring, your new text, and then the final substring.

From the command line you could do

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cmd /v:on
set var=asdfg
echo !var:~0,1!OK!!var:~2!

Normally you would do something like this from a batch file, in which case you can use SETLOCAL to enable delayed expansion. Also you need to escape the ! character in a batch file whenever delayed expansion is enabled so that cmd.exe does not attempt to expand it. If enclosed within quotes then you can escape it simply as "^!". But if not within quotes then you must escape it as ^^^! Finally, if you are setting a variable to the new value, then you can enclose the entire expression in quotes and use the simpler escape sequence, yet the quotes won't be included in the value.

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@echo off
setlocal enableDelayedExpansion
set var=asdf
echo "!var:~0,1!OK^!!var:~2!"
echo !var:~0,1!OK^^^!!var:~2!
set "var=!var:~0,1!OK^^^!!var:~2!"
set var

Since your sample string does not contain any special characters, you don't need delayed expansion. In this case you don't need to escape the ! Note that I explicitly disabled delayed expansion, though normally it is the default state.

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@echo off
setlocal disableDelayedExpansion
set var=asdf
echo %var:~0,1%OK!%var:~2%
set var=%var:~0,1%OK!%var:~2%
set var

Dave Benham

[Squashman]

I think this is what you were actually trying to do.

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@echo off
setlocal enabledelayedexpansion
set var=asdfg
echo !var:%var:~1,1%=OK!

output

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E:\batch files\SET_example>example.bat
aOKdfg

EDIT: Dave Beat me!!!

[dbenham]

Ahh - but you figured out what atamo was attempting to do with the original syntax. I missed that.

However, if the 2nd character happens to appear in multiple locations, it will be replaced in all of them with your fixed version I'm not sure if that is what is actually wanted.


Dave Benham

[atamo]

thank's your fast help

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echo !var:%var:~1,1%=OK!  work nice that I want, get   aOKdfg

but in this case why !var:s=OK! don't expanded ?

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echo !var:!var:~1,1!=OK!    don't work, I get       !var:s=OK! 

[Aacini]

My SET/S macro do exactly what you need (and much more):

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%SET/S% variable:~start,size=substring to insert

Previous "SET-Substring" macro replace part of the current "variable" value with another substring; the part to replace is given in the same way as the standard Batch substring extraction with the sole difference that it does not end at the last character by default, so this point must be explicitly indicated by -0 in "size".

For example:

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REM Insert the SET/S macro definition here...

set var=asdfg
%set/s% var:~1,1=OK
echo %var%

aOKdfg

You must note that your original code does NOT replace by position, but replaces all instances of the s letter in the string by the word OK.