Match all "http" only URLs without additional characters

I have tried the below expressions.

(http:\/\/.*?)['\开发者_开发知识库"\< \>]


(http:\/\/[-a-zA-Z0-9+&@#\/%?=~_|!:,.;\"]*[-a-zA-Z0-9+&@#\/%=~_|\"])

The first one is doing well but always gives the last extra character with the matched URLs.

Eg:

http://domain.com/path.html" 

http://domain.com/path.html<

Notice

" <

I don't want them with URLs.

You can use lookahead instead of making ['\"\< >] part of your match, i.e.:

(http:\/\/.*?)(?=['\"\< >])

Generally speaking, whereas ab matches ab, a(?=b) matches a (if it's followed by b).

References

• regular-expressions.info/Lookarounds

Related questions

• How does the regular expression (?<=#)[^#]+(?=#) work?

Capturing group option

Lookarounds are not supported by all flavors. More widely supported are capturing groups.

Generally speaking, whereas (a)b still matches ab, it also captures a in group 1.

References

• regular-expressions.info/Round Brackets for Grouping

Related questions

• How can I match on, but exclude a regex pattern?

Negated character class option

Depending on the need, often times using a negated character class is much better than using a reluctant .*? (followed by a lookahead to assert the terminator pattern in this case).

Let's consider the problem of matching "everything between A and ZZ". As it turns out, this specification is ambiguous: we will come up with 3 patterns that does this, and they will yield different matches. Which one is "correct" depends on the expectation, which is not properly conveyed in the original statement.

We use the following as input:

eeAiiZooAuuZZeeeZZfff

We use 3 different patterns:

• A(.*)ZZ yields 1 match: AiiZooAuuZZeeeZZ (as seen on ideone.com)
  • This is the greedy variant; group 1 matched and captured iiZooAuuZZeee
• A(.*?)ZZ yields 1 match: AiiZooAuuZZ (as seen on ideone.com)
  • This is the reluctant variant; group 1 matched and captured iiZooAuu
• A([^Z]*)ZZ yields 1 match: AuuZZ (as seen on ideone.com)
  • This is the negated character class variant; group 1 matched and captured uu

Here's a visual representation of what they matched:

         ___n
        /   \              n = negated character class
eeAiiZooAuuZZeeeZZfff      r = reluctant
  \_________/r   /         g = greedy
   \____________/g

References

• regular-expressions.info/Character Class and Repetition: An Alternative to Laziness

Related questions

• Difference between .*? and .* for regex

You need to use "(?=regex)" (lookahead), which lookups a particular pattern, but doesn't include it in the result:

http:\/\/.*?(?=['\"\< >])

Hmmm, I'd probably do this simply by saying "keep going until you get an unwanted character", like so:

http://[^'"< >]*

Escaped version (based on Q - not sure what engine this is):

http:\/\/[^'\"\< >]*

However the lookahead solution by polygenelubricants is a more flexible way, if you might have some of those characters in the URL (but not at the end).