How to detect if input is quote?

I have the following code

if "%userInput%"==""" (
    do_something
)

I would like it to detect if the %userInput% is a quote("). However, this code throws an error.

How to detect if input is a quote?

回答1:

Here is a solution without delayed expansion that should work with any input, including spaces and poison characters. My test code is in a loop. When you are ready to quit, simply press <Enter> without typing anything.

@echo off
setlocal

:loop
set "var="
set /p "var=enter a string: "
if not defined var exit /b

if "%var:"=""%" == """" echo equals quote

echo(
goto loop

The trick should be fairly obvious. Enclose the value in quotes, and double all internal quotes during expansion, such that all token delimiters and poison characters are guaranteed to be quoted. Of course on the right side you must include the enclosing quotes, plus the doubled quote you are trying to match (for a total of 4).

If you need to be able to pass all input (including no value) through the IF test, then you can define a test variable. Something like the following (assume var already has the user input)

set "test="
if defined var set "test=%var:"=""%"
if "%test%" == """" echo equals quote

Alternate solution using FINDSTR

I was inspired by Stephan's answer to find an alternative solution using FINDSTR that does not require delayed expansion. He interpreted the question differently, answering the question "Does the input value contain a quote anywhere within it?", and solved that using FIND.

But FIND cannot determine if the input exactly matches a single quote character.

A simple one liner with FINDSTR can solve this.

2>nul set var|findstr /x ^"var=\"" >nul && echo equals quote

2>nul set var|findstr /x "var=\"^" >nul && echo equals quote

Note that FINDSTR generally requires quotes to be escaped as \", and the batch parser requires one of the quotes to be escaped as ^" because there are an odd number of quotes. If a quote is not escaped, then the remainder of the line is considered to be part of the string instead of redirection and conditional execution operations.

Also note that with the batch parser, it is impossible to escape a quote once quoting is activated. So the following does NOT work

2>nul set var|findstr /x "var=\^"" >nul && echo equals quote THIS DOES NOT WORK

回答2:

as you asked "if input contains a quote". Just find it:

set "i=hello " world"
set i|find """" >nul && echo yes || echo no

Note: echo %i%|find """" may not work with unpaired quotes, but set doesn't care.

Note the """" syntax (looks strange and is unintuitive). find doesn't seem to use the usual escaping: find "^"" gives syntax error.

回答3:

with additional variable and delayed expansion:

@echo off

set /p quote=Enter quote
set "test_quote=""
setlocal enableDelayedExpansion

if !quote! equ !test_quote! echo equal

回答4:

…and possibly without delayed expansion:

@Echo Off
Set/P "input=Please input a character: "
If [^%input%]==[^"] (Echo it is a quote) Else Echo it is something else
Timeout -1