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How do I match a quoted string within quotes?

/^"((?:[^"]|\\.)*)"/

Against this line:

"quote\_with\\escaped\"characters" more

It corresponds only \"

, even though I clearly defined \

as the escape-character (and it corresponds to \_

, and \\

fine ...).

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3 answers


It works correctly if you reverse the order of your two alternatives:

/^"((?:\\.|[^"])*)"/


The problem is that otherwise the important character \

gets eaten before it tries to match \"

. It worked before for \\

and \_

only because both characters in each pair match yours [^"]

.

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Using Python with source string literals to ensure no further interpretation of escape sequences works, the following option works:

import re

x = re.compile(r'^"((?:[^"\\]|\\.)*)"')

s = r'"quote\_with\\escaped\"characters" more"'

mo = x.match(s)
print mo.group()


emits "quote\_with\\escaped\"characters"

; I believe that in your version (which will also abort the game early if replaced here) the subexpression "not double" ( [^"]

) will swallow the backslashes you intend to take as escaping immediately following characters. All I'm doing here is that backslashes like this are NOT swallowed this way, and as I said, they seem to work with this change.

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Not going to be confused, just different information that I played with. Below regexp (PCRE) try not to match incorrect syntax (eg end with \ ") and can be used with either" or "

/('|").*\\\1.*?[^\\]\1/

for use with php

<?php if (preg_match('/(\'|").*\\\\\1.*?[^\\\\]\1/', $subject)) return true; ?>

For:

"quote\_with\\escaped\"characters"  "aaa"
'just \'another\' quote "example\"'
"Wrong syntax \"
"No escapes, no match here"

This is the only match:

"quote\_with\\escaped\"characters" and
'just \'another\' quote "example\"'
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