Re: Re: Match anything between two " that is not a " except if it is escaped...

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)