Regex for URL with query string

Question

I'm trying to create a regular expression to use as a rule in fiddler. I'm not very good at regular expressions.

This regular expression:

http://myServer:28020/MyService/ItemWebService.json?([a-zA-Z]+)

Matches the URL below:

http://myServer:28020/MyService/ItemWebService.json?action=keywordSearch&username=StockOnHandPortlet&sessionId=2H7Rr9kCWPgIZfrxQiDHKp0&keywords=blue&itemStatus=A

So far so good. But why when I try this regular expression:

http://myServer:28020/MyService/ItemWebService.json?action=keywordSearch([a-zA-Z]+)

It does not match the above URL. Why would that be?

tried that. It doesn't work either. Does the & need to be escaped? — Richie
– Richie, Commented Jul 12, 2015 at 11:36
What do you need to match? The whole URL? Only the query string? Or the part before the query string? Mind that . and ? must be escaped outside of character class. — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Jul 12, 2015 at 11:43
the whole url. Before fiddler intercepts the http requests it wants to check that a url matches the pattern specified. — Richie
– Richie, Commented Jul 12, 2015 at 11:47

Wiktor Stribiżew · Accepted Answer · 2015-07-12 12:17:28Z

11

In a regular expression, you need to escape . and ? outside of character class.

So, this is enough to match the URL itself:

http://myServer:28020/MyService/ItemWebService\.json

URL with query string can be matched with

http://myServer:28020/MyService/ItemWebService\.json\??(?:&?[^=&]*=[^=&]*)*

See demo

Explanation:

http://myServer:28020/MyService/ItemWebService\.json - matches the base URL where we need to escape . to match it literally
\?? - matches 0 or 1 (due to ? quantifier) literal ?
(?:&?[^=&]*=[^=&]*)* - matches 0 or more (due to *) sequences of &?[^=&]*=[^=&]*:
- &? - 0 or 1 & (no need escaping)
- [^=&]* - 0 or more characters other than = and &
- = - n equals sign
- [^=&]* - 0 or more characters other than = and &

If you want to match a URL that has the first query parameter=value set to action=keywordSearch, you can use the following regex:

http://myServer:28020/MyService/ItemWebService\.json\?action=keywordSearch(?:&?[^=&]*=[^=&]*)*

edited Jul 12, 2015 at 12:17

answered Jul 12, 2015 at 11:48

Wiktor Stribiżew

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3 Comments

Richie Over a year ago

thanks. This worked for me. I understand your explanation as well about escaping. Can you elaborate on the bit inside the brackets for us beginners.

Richie Over a year ago

thanks. Actually sorry to bother but could you modify you answer to include the action=keywordSearch. I was always able to match the url without the action part but it was when i added the action part that i couldn't get the regex right.

Wiktor Stribiżew Over a year ago

Ok, I added that version.

BhavO · Accepted Answer · 2015-07-12 11:51:27Z

2

Use

(\?|\&)([^=]+)\=([^&]+)

With g option (global)

http://myServer:28020/MyService/ItemWebService.json(\?|\&)([^=]+)\=([^&]+)

edited Jul 12, 2015 at 11:51

answered Jul 12, 2015 at 11:38

BhavO

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3 Comments

Richie Over a year ago

Sorry but fiddler also said that the url does not match the pattern

BhavO Over a year ago

Add your URL in front of it

Himanshu Bansal Over a year ago

just using ([^=]+)\=([^&]+) matching asd and asd=asd and also asd=.... how do I use this with pattern attribute to match only the query string?

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