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[Batch/CMD] - Does variable contain numbers, spaces, one word?


haze1434
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Hi all,

I have a batch file which starts with the user inputting either;

  • The hostname for a PC
  • The full name of a user
  • The surname of a user

I would like the script to work out what was input and GOTO the next relevant section.

Basically;

Does the variable contain any numbers? If yes, GOTO Hostname

If not;

Does the variable contain any spaces? If yes, GOTO FullName

If not;

Does the variable contain only 1 word made up of only letters? If yes, GOTO Surname

If not;

Error message.

I have tried various instances of...

echo %INPUT%|findstr /r "[^a-zA-Z]" > nul

... and similar, but I can't seem to get it work correctly.

Thanks in advance.

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So the current code I have is...

echo %StartInput% | findstr /m /r "[0-9]"
IF ERRORLEVEL 1 ECHO This includes numbers
echo %StartInput% | findstr /m /r "[^a-zA-Z]"
IF ERRORLEVEL 1 ECHO This doesn't include numbers
Pause

... but the output, when I enter '1234' as the variable %StartInput% is...

1234
1234
Press any key to continue . . .

It appears it's ignoring the findstrs all together and is simply echoing the varibale on-screen twice.

Using your example doesn't work either, unfortunately, as it does the same...

echo "%StartInput%" | findstr /m /r "[0-9]"
IF %ERRORLEVEL% EQU 1 echo "Numbers"
echo "%StartInput%" | findstr /m /r "[^a-zA-Z]"
IF %ERRORLEVEL% EQU 1 echo "Letters Only"
Pause

... and the output, when I enter '1234' as the variable %StartInput% is...

"1234"
"1234"
Press any key to continue . . .

I've also tried;

echo "%StartInput%" | findstr /m /r "[^0-9]"
echo "%StartInput%"|findstr /m /r "[0-9]"
echo "%StartInput%"|findstr "[0-9]"

All give the same output, namely repeating the variable twice.

Hmmm!

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Did you try to echo ERRORLEVEL itself? It might be that it doesn't get set due to the pipe command, so try to echo the value into a file, then pass the filename as a parameter to findstr.

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A while back I found a batch script for OS version detection, it's doing something similar, maybe you can adapt it:

SET OSVersion=Unknown

VER | FINDSTR /L "5.0" > NUL

IF %ERRORLEVEL% EQU 0 SET OSVersion=2000

VER | FINDSTR /L "5.1." > NUL

IF %ERRORLEVEL% EQU 0 SET OSVersion=XP

VER | FINDSTR /L "5.2." > NUL

IF %ERRORLEVEL% EQU 0 SET OSVersion=2003

VER | FINDSTR /L "6.0." > NUL

IF %ERRORLEVEL% EQU 0 SET OSVersion=Vista

VER | FINDSTR /L "6.1." > NUL

IF %ERRORLEVEL% EQU 0 SET OSVersion=7

VER | FINDSTR /L "6.2." > NUL

IF %ERRORLEVEL% EQU 0 SET OSVersion=8

IF %OSVersion%==Unknown (

ECHO Unable to determine your version of Windows.

) ELSE (

ECHO You appear to be using Windows %OSVersion%

)

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