I need to have a batch file that checks if a variable has a quote character for the first character. I have tried this:
set fname=%1
set bQuote=0
if %fname:~0,1% == ^" (
set bQuote=1
)

When I run this, I get an ERROR '( was unexpected at this time.' I assume it's because the quote character isn't escaped properly. When there is a quote character, the line reads: if " == ^", which leaves the first quote as 'open'. Without the ^ doesn't work either.

Any ideas?What is the purpose (briefly)? Is the problem that the variable could be either unquoted, or quoted at both beginning and END, or liable to just have a leading quote?

You do know about the tilde (~) MODIFIER, which strips leading and/or surrounding quotes (if PRESENT)? It does NOTHING if there are no quotes.

i.e. if %1 is "hello world", or "hello world then %~1 becomes hello world, but if %1 is hello world then %~1 is also hello world





Any other language, a regular expression (RegEx) would be the way to go. In batch, delayed expansion should help out:

Code: [Select]@echo off
setlocal enabledelayedexpansion

set fname=%1
set quote="
set bquote=0

if "!fname:~0,1!" EQU "!quote!" (
set bquote=1
)

Good luck Thank you all. I think I've got it solved.Quote from: Sidewinder on February 05, 2014, 07:53:55 AM

Any other language, a regular expression (RegEx) would be the way to go.