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    <pre><code>/(?&lt;=&amp;|\?)foo(=[^&amp;]*)?(&amp;|$)/ </code></pre> <p>Uses lookbehind and the last group to "anchor" the match, and allows a missing value. Change the <code>\?</code> to <code>^</code> if you've already stripped off the question mark from the query string.</p> <p>Regex is still not a substitute for a real parser of the query string, however.</p> <p><em>Update:</em> Test script: (run it at <a href="http://codepad.org/" rel="nofollow noreferrer">codepad.org</a>)</p> <pre><code>import re regex = r"(^|(?&lt;=&amp;))foo(=[^&amp;]*)?(&amp;|$)" cases = { "foo=123": "", "foo=123&amp;bar=456": "bar=456", "bar=456&amp;foo=123": "bar=456", "abc=789&amp;foo=123&amp;bar=456": "abc=789&amp;bar=456", "oopsfoo=123": "oopsfoo=123", "oopsfoo=123&amp;bar=456": "oopsfoo=123&amp;bar=456", "bar=456&amp;oopsfoo=123": "bar=456&amp;oopsfoo=123", "abc=789&amp;oopsfoo=123&amp;bar=456": "abc=789&amp;oopsfoo=123&amp;bar=456", "foo": "", "foo&amp;bar=456": "bar=456", "bar=456&amp;foo": "bar=456", "abc=789&amp;foo&amp;bar=456": "abc=789&amp;bar=456", "foo=": "", "foo=&amp;bar=456": "bar=456", "bar=456&amp;foo=": "bar=456", "abc=789&amp;foo=&amp;bar=456": "abc=789&amp;bar=456", } failures = 0 for input, expected in cases.items(): got = re.sub(regex, "", input) if got != expected: print "failed: input=%r expected=%r got=%r" % (input, expected, got) failures += 1 if not failures: print "Success" </code></pre> <p>It shows where my approach failed, Mark has the right of it&mdash;which should show why you shouldn't do this with regex.. :P</p> <hr> <p>The problem is associating the query parameter with exactly one ampersand, and&mdash;if you must use regex (if you haven't picked up on it :P, I'd use a separate parser, which might use regex inside it, but still actually understand the format)&mdash;one solution would be to make sure there's exactly one ampersand per parameter: replace the leading <code>?</code> with a <code>&amp;</code>.</p> <p>This gives <code>/&amp;foo(=[^&amp;]*)?(?=&amp;|$)/</code>, which is very straight forward and the best you're going to get. Remove the leading <code>&amp;</code> in the final result (or change it back into a <code>?</code>, etc.). Modifying the test case to do this uses the same cases as above, and changes the loop to:</p> <pre><code>failures = 0 for input, expected in cases.items(): input = "&amp;" + input got = re.sub(regex, "", input) if got[:1] == "&amp;": got = got[1:] if got != expected: print "failed: input=%r expected=%r got=%r" % (input, expected, got) failures += 1 if not failures: print "Success" </code></pre>
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    1. COhaving some problems with this one, but i'm working on it. yes, there is no \?, my string is only the query string
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    2. COfails for these inputs: `bar=456&foo`, `bar=456&foo=`, `bar=456&foo=123`
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    3. COYes, I know, that's why I said my approach fails. :)
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