Need help with regex replace in php

Question

i have a string which includes links of this pattern:

<a href="http://randomurl.com/random_string;url=http://anotherrandomurl.com/">xxxx</a>

i want to remove "http://xxx.xxx.xxx/random_string;url=" and keep the rest of the string, leaving at the end

<a href="http://anotherrandomurl.com/">xxxx</a>

Can anyone help please ?

You have not posted any code or previous attempt, and we can't possibly entertain one question per need of a regular expression. If you can edit your question to show the regex you tried, it will gain answers that actually _help people understand regular expressions_. Please flag your question for moderator attention if you do. — Tim Post, Aug 25 '11 at 15:49

score 1 · Answer 1 · answered Aug 25 '11 at 14:45

1

Use:

$new_link = preg_replace('/<a href="(?:.+);url=([^"]+)">/', '<a href="$1">', $url);

answered Aug 25 '11 at 14:45

piotrp

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`?:` makes this a non-capturing subpattern – piotrp Aug 25 '11 at 14:49
This will fail if the tag has any other attributes, or different amounts of whitespace, or if the URL is quoted with `'` instead of `"`. – Justin Morgan - On strike Aug 25 '11 at 14:56

score 1 · Answer 2 · answered Aug 25 '11 at 14:47

1

There are multiple methods for achieving your desired result. An alternative from regex would be to find the occurence of url= using strpos and remove those characters and the preceeding characters as well.

answered Aug 25 '11 at 14:47

Robert

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+1 for a solution that will be more robust and easier than a regex – afuzzyllama Aug 25 '11 at 14:58

score 1 · Answer 3 · edited May 23 '17 at 11:47

This is trickier than you think, and I urge you to avoid using regex for it.

Instead, you should use an HTML parser to find all <a> tags in the document, then split their href attributes on ;url= and keep only the last part.

However, if you must use a regex, the following should work for most well-formed HTML:

preg_replace('/(<\s*a\s[^>]*href=)(["\'])(?:[^\1]*;url=)([^\1]*)(\1[^>]*>)/i', "$1$2$3$4", $url)

Explanation:

(<\s*a\s[^>]*\bhref=) # <a, optionally followed by other attributes, and then href. Whitespace is ignored. This will be captured in backreference $1.
(["\'])               # Either " or ' to enclose the href value. This will be captured in $2 for matching later.
(?:[^\1]*;url=)       # Any number of URLs followed by ";url=". This will be thrown out.
([^\1]*)              # This is the URL you want to keep. It will keep matching until the end of the quotes. This will be captured into $3.
(\1[^>]*>)            # The remainder of the <a> tag, including any other attributes. This is captured in $4.

score 0 · Answer 4 · answered Aug 25 '11 at 14:50

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$new_link = preg_replace('~(\shref=")[^"]+?(?<=;url=)~', '$1', $url);

answered Aug 25 '11 at 14:50

Geert

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Need help with regex replace in php

4 Answers4

This is trickier than you think, and I urge you to avoid using regex for it.