Regular Expression help to match below code

Question

I want a regular expression to fetch any URL between double quotes.

<a href="http://www.any-web_address.com">
<a href="http://142.12.64.71:8083">

score 0 · Answer 1 · edited Jul 04 '16 at 09:07

0

"http://[0-9 a-z A-Z . : ]{1,100}"

edited Jul 04 '16 at 09:07

F.P

answered Jul 04 '16 at 08:32

user561810

1

Although this code may be help to solve the problem, providing additional context regarding _why_ and/or _how_ it answers the question would significantly improve its long-term value. Please [edit] your answer to add some explanation. In particular, I don't see how this matches URLs such as `ftp://server/file` or `mailto:user@host`... – Toby Speight Jul 04 '16 at 16:46

Michał M · Answer 2 · 2016-07-04T09:14:48.260

0

Something like this?

\"\K([\w\:\/\.\-]+)

If You want with double quotes (it was said "fetch any url between double quotes" so I thought without \"):

\"([\w\:\/\.\-]+)\"

edited Jul 04 '16 at 09:14

answered Jul 04 '16 at 08:33

Michał M

Side question: Why the `\K` if you match with a group anyway? Just curious. – F.P Jul 04 '16 at 09:07
It was said "fetch any url BETWEEN double quotes" so in my understanding - without them. So find first \", clear position - \K and then rest of regex - match url. – Michał M Jul 04 '16 at 09:13

Maria Ivanova · Answer 3 · 2016-07-04T09:07:23.030

0

Here is my suggestion (in case your regex flavour supports lookarounds):

(?<=href="|link="|src=")(((http|https)(:\/\/))?([\/\w\-]{2,})(([\.])([\w\-]*)){1,})([\w.,@?^=%&amp;:\/~+#-]*[\w@?^=%&amp;\/~+#-]*)(?=")

edited Jul 04 '16 at 09:07

answered Jul 04 '16 at 08:37

Maria Ivanova

suppose I have something like this in code , It will select that too – user561810 Jul 04 '16 at 08:50
I've made an edit on the regex. Now it will only match after `href="` or `link="` – Maria Ivanova Jul 04 '16 at 08:54

score 0 · Answer 4 · edited May 23 '17 at 11:58

0

If you do not reduce the scope of your problem, this post : Why it's not possible to use regex to parse HTML/XML: a formal explanation in layman's terms may help you. Else, for instance if you only want the URIs after href=, you can do this like that :

/(?:href=")(.[^"]*)"/g

edited May 23 '17 at 11:58

Community

answered Jul 04 '16 at 08:40

M. Timtow

it will include keyword href=" also P.S:- I'm new to regex so might be wrong. – user561810 Jul 04 '16 at 08:48
1

(?:blablabla...) is a non-capturing group (because of ?:), whereas (blablabla...) is a capturing one, so no it won't. You can test your regex here : https://regex101.com/#javascript it explains pretty well. – M. Timtow Jul 04 '16 at 08:52
http(.[^"]*) this will also help – user561810 Jul 04 '16 at 08:52
Yes, you are right, but this will not capture http then. – M. Timtow Jul 04 '16 at 08:53

CinCout · Answer 5 · 2016-07-04T09:42:44.550

0

Use the following regex:

.*href="(\S*)" demo

We are effectively looking for the presence of href= and then capturing all non-whitespace characters that appear between double quotes.

edited Jul 04 '16 at 09:42

answered Jul 04 '16 at 09:01

CinCout

5 Answers5