Findstr question

[MrKnowItAllxx]

Maybe I am missing something simple, but I would like to use findstr to search for a line in a file, and parse that line to give output (a for loop works). I have that down, but I would also like the script to alert the user if no match is found.

So far I have this:

for /f "delims=." %%f in ('findstr /i /c:"%%d" "file.txt"') do echo [%%f]

By "parse" the line I mean that I would like to take only the text before the first "." in the file, but again I need to know if the line wasn't found in the first place.

[Ed Dyreen]

'
only the text before the first "." if the line was found.

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for /f "tokens=1 delims=." %%f in ( '2^>nul findstr /lic:"%%~d" "file.txt"' ) do echo.[%%f]

[MrKnowItAllxx]

Your code works just fine, but what I need is for there to be a message printed on the screen if the search string (%%d) is not found in the file.

[Ed Dyreen]

'

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@echo off &setlocal enableDelayedExpansion

set "$file=file.txt"
set  "$str=something"

2>&1 >nul findstr /lic:"!$str!" "!$file!" &&echo.found !$str! in file. ||echo.not found !$str! in file.

for /f "tokens=1 delims=." %%? in (

       '2^>^&1 ^>nul findstr /lic:"!$str!" "!$file!" ^&^&echo.found !$str! in file. ^|^|echo.not found !$str! in file.'

) do   echo.%%?

pause
exit

'file.txt'

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something

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found something in file.
found something in file
Druk op een toets om door te gaan. . .

'file.txt'

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nothing

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not found something in file.
not found something in file
Druk op een toets om door te gaan. . .