batch script, find string in text file by line then replace whole line with another string

Question

I want to write a batch file that find a given string in a text file by line then replace whole that line with another given string.

example : has file text.txt string_replace = abc string_replace_with = xyz if every line in text.txt that contains string_replace, so that line will be replaced with string_replace_with

Can any one help me ? Thanks alot

Answer 1

@echo off
SET InFile=test.txt
FOR /F "tokens=*" %%A IN ('FINDSTR "xyz" "%InFile%"') DO CALL :FindString "%%A"
pause
goto :eof
:FindString
SET String=%~1
SET String=%String:xyz=abc%
echo.%String%>getout.txt

Try these if it solve your problem. From your question what I understood is that you want to find and replace the word with desired word.

Answer 2

为自己准备一个Windows sed端口。

Answer 3

If you launch this SAR.BAT file like this it will replace abc with xyz

SAR "fileinput.txt" "fileoutput.txt" "abc" "xyz"

::Search and replace
@echo off
if "%~3"=="" (
echo.Search and replace
echo Syntax: 
echo %0 "filein.txt" "fileout.ext" "regex" "replace_text" [first]
echo.
echo. if [first] is present only the first occurrence is changed
goto :EOF
)
if "%~5"=="" (set global=true) else (set global=false)
set s=regex.replace(wscript.stdin.readall,"%~4")
 >_.vbs echo set regex=new regexp
>>_.vbs echo regex.global=%global%
>>_.vbs echo regEx.IgnoreCase=True            
>>_.vbs echo regex.pattern="%~3"
>>_.vbs echo wscript.stdOut.write %s%
cscript /nologo _.vbs <"%~1" >"%~2"
del _.vbs