Max Antonov schreef:
mathieu leddet writes:
Hi everyone,
I am struggling with regular expression trying to match strings
delimited by double quotes, but taking into consideration that \" is not
a string ending character.
......
// pattern for catching strings between "
$pattern = '#"([^"]*)"#';
.....
$out contains : this is a string : <b>"Hi everyone my name is
\"</b>Mathieu\<b>"!"</b>
Here is a second string : <b>"PHP is just perfect"</b>
.....
--
Mathieu
If I right understand you scope. (yes, my English is bad)
better than our russian.
You hope to get result such as:
this is a string : <b>"Hi everyone my name is \"Mathieu\"!"</b> Here is
a second string : <b>"PHP is just perfect"</b>
I try to fix you regular expression.
$pattern = '#"(.*?)(?<=[^\\\\])"#is';
attend to this: (?<=[^\\\\])
attend? don't understand what you mean BUT you have given the OP the
answer by changing his regexp to include a [negative] look behind assertion
for the backslash. :-)
when PHP compile this string - inside it looks like (?<=[^\\])
when regular expression compile inside pcre library it looks like
#"(.*?)(?<=[^\])"#is
this part: (?<=[^\])" is mean folow:
double quote, which not have leading backslash.
see folow:
max@maximus:~$ cat preg.php
<?php
$in = 'this is a string : "Hi everyone my name is \"Mathieu\"!" Here is
a second string : "PHP is just perfect"';
// pattern for catching strings between "
$pattern = '#"(.*?)(?<=[^\\\\])"#is';
// surround matching string with HTML span code to highlight
$replacement = '<b>"${1}"</b>';
// perform the reg exp replacement
$out = preg_replace($pattern, $replacement, $in);
echo $out,"\n\n";
max@maximus:~$ php preg.php
this is a string : <b>"Hi everyone my name is \"Mathieu\"!"</b> Here is
a second string : <b>"PHP is just perfect"</b>
Is this rigth?
--
Max Anotnov (idler at instanceof dot ru)
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