@echo off
SET /p var=Enter:
echo %var% | findstr /r "^[a-z]{2,3}$">nul 2>&1
if errorlevel 1 (echo does not contain) else (echo contains)
pause
I'm trying to valid a input which should contain 2 or 3 letters. But I tried all the possible answer, it only runs if error level 1 (echo does not contain).
Can someone help me please. thanks a lot.
findstr has no full REGEX Support. Especially no {Count}. You have to use a workaround:
echo %var%|findstr /r "^[a-z][a-z]$ ^[a-z][a-z][a-z]$"
which searches for ^[a-z][a-z]$ OR ^[a-z][a-z][a-z]$
(Note: there is no space between %var% and | - it would be part of the string)
Since other answers are not against findstr, howabout running cscript? This allows us to use a proper Javascript regex engine.
@echo off
SET /p var=Enter:
cscript //nologo match.js "^[a-z]{2,3}$" "%var%"
if errorlevel 1 (echo does not contain) else (echo contains)
pause
Where match.js is defined as:
if (WScript.Arguments.Count() !== 2) {
WScript.Echo("Syntax: match.js regex string");
WScript.Quit(1);
}
var rx = new RegExp(WScript.Arguments(0), "i");
var str = WScript.Arguments(1);
WScript.Quit(str.match(rx) ? 0 : 1);
Stephan's answer is correct in terms of support for regular expression. However, it does not regard a bug of findstr concerning character classes like [a-z] -- see this answer by dbenham.
To overcome this you need to specify this ( I know it looks terrible):
echo %var%|findstr /R "^[abcdefghijklmnopqrstuvwxyz][abcdefghijklmnopqrstuvwxyz]$ ^[abcdefghijklmnopqrstuvwxyz][abcdefghijklmnopqrstuvwxyz][abcdefghijklmnopqrstuvwxyz]$"
This truly matches only strings consisting of two or three lower-case letters. Using ranges [a-z] would match lower- and upper-case letters except Z.
For a complete list of bugs and features of findstr, reference this post by dbenham.
errorlevel is that number OR HIGHER.
Use following.
if errorlevel 1 if not errorlevel 2 echo It's just one.
See this
Microsoft Windows [Version 10.0.10240]
(c) 2015 Microsoft Corporation. All rights reserved.
C:\Windows\system32>if errorlevel 1 if not errorlevel 2 echo It's just one.
C:\Windows\system32>if errorlevel 0 if not errorlevel 1 echo It's just ohh.
It's just ohh.
C:\Windows\system32>
If Higher than one and not higher than n+1 (2 in this case)
findstr errorlevel doesn't give the count, but only 0 (found at least one) or 1 (found none).