Batch replace string without a FOR loop

问题:

I have the following string

set str=aaaaa,bbbbb,ccccc,dddd

I need to replace everything from the first , (comma) till the end of string. but I don't know how the string starts.

I can't use the * wildcard as it is not in the start of string, so something like that -

set str=%str:,*=% >>>> Will not work

Any ideas ? it must be in a batch script

回答1:

There is a very simple and fast solution that avoids a FOR loop: use search and replace to inject a concatenated REM command at each comma :-)

@echo off
set str=aaaaa,bbbbb,ccccc,dddd
set "str=%str:,="&rem %
echo str=%str%

Explanation

Here is the the important line of code, before any expansion takes place:

set "str=%str:,="&rem %

And here is what the line is transformed into after variable expansion:

set "str=aaaaa"&rem bbbbb"&rem ccccc"&rem dddd

It becomes a SET statement, followed by a REM statement. The & operator allows you to combine multiple commands on one line. There is only one REM statement because everything after the first REM is considered part of the remark.

回答2:

To remove everything after the first comma you can use

@echo off
setlocal enabledelayedexpansion
set str=aaaaa,bbbbb,ccccc,dddd
for /f "delims=," %%a in ("%str%") do (
set str=%%a
echo !str!
)
pause >nul

To replace, simply strip it and append the new string on the end

    set str=%%a,new,string

Update

This will do the same, but without a for loop

@echo off
setlocal enabledelayedexpansion
set str=aaaaa,bbbbb,ccccc,dddd
set strip=%str%
set length=0
set start=0
set new=
:loop
if defined strip (
set strip=!strip:~1!
set /a length+=1
goto :loop
)
:comma
if %length% gtr 0 (
call :add !start!
if "!char!"=="," (
goto :break
) else (
set new=!new!!char!
set /a length-=1
set /a start+=1
goto comma
)
)

:break
echo %new%
pause >nul

:add
set char=!str:~%1,1!