perl正则表达式

Question

I am having some URLs in this format. Some URLs contain &abc=4 and some not.

xxxxxxxxxxxxxxxxxxxxxxxxxxx&abc=4
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx&abc=4
xxxxxxxxxxxxxxxxxxxxx
xxxxxxxxxxxxxxxxxxxxxxxxxxxxx

here xxxxxxxxxxxxxxxxxxxxx is string

I want to match URLs which have xxxxxxxxxxxxxxxxx only and not &abc=4 (meaning I want to get these type of URLs, only xxxxxxxxxxxxxx , xxxxxxxxxxxxxx , xxx )

I know how to write a regular expression which matches the entire url. For example: /x.*abc=4/

But how do I write a regular expression that matches only xxxxxxxxxx and not &abc=4 ?

Answer 1

I would use negative look-ahead assertion (Look ahead what is not allowed to follow my pattern)

^(?!.*&abc=4$).*$

This pattern will match any string that does not end with &abc=4

you can verify it online here: http://www.rubular.com/

Answer 2

Use negative lookbehind assertion . The form is:

(?<![&?]abc=4)

(this will also exclude ?abc=4 ).

Answer 3

Assuming your URLs are on each line, you can use:

([^&]+?)

This basically will match anything up to the the first instance of &.

As @Benoit said, you can do this using a zero width expression to negate the capture of the query string, but you would be after a positive lookahead, and not a negative lookbehind, syntax example below:

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

As you can see though, this would complicate the expression a touch.

Hope this helps.