replace a pattern to extract an image url from data php

Question

I am trying to figure it out how to extract the url from this pattern

![enter image description here][1] [1]: http://www.codefixup.com/wp-content/uploads/2016/09/pattern-matching-in-php.png

I just need the http part so I can place it inside an image tag. The image description could change, how can I do it with regex or preg_replace?

Do you only want canonical URLs? I.E you only want URLs that start with http or do you also want URLs like www.google.com — bassxzero
– bassxzero, Commented Sep 19, 2016 at 16:19
I need the whole after "[1]:", the http://...someurl until space since te pattern above will be inside a text. — jose
– jose, Commented Sep 19, 2016 at 16:26
Example: Hi boy, how are you? This is my cat ![enter image description here][1] [1]: somewebsite.com/images/mycat.jpg I hope you love it. — jose
– jose, Commented Sep 19, 2016 at 16:28
@jose -- I added an answer that should get what you need, assuming the format is always the same — A_Elric
– A_Elric, Commented Sep 19, 2016 at 16:30

A_Elric · Accepted Answer · 2016-09-19 17:52:53Z

1

Assuming that you wanted to capture the entire url the following should work

(?:\: )(?P<URL>.*)

Example: https://regex101.com/r/aV1fJ7/1

Note this would NOT work if you had a description that was like "Check out this cool dog: "

Or something along those lines.

I can make this more specific by doing...

(?:\[\d+]\: )(?P<URL>.*)

Live example of this working with more specific one: https://regex101.com/r/aV1fJ7/2

Example 3 to pull from an img src="" tag

(?:src=\")(?P<URL>.*)\"

https://regex101.com/r/aV1fJ7/3

And some example code showing how to capture, transform and output an image tag:

<?php

$urlstring ='![enter image description here][1] [1]: http://www.codefixup.com/wp-content/uploads/2016/09/pattern-matching-in-php.png';
$regex = '/ (?:\[\d+\]\: )(?P<URL>.*)/';
//echo $regex.'--'.$urlstring;
if (preg_match($regex, $urlstring, $matches)) {
    echo "img src=\"".$matches[1]."\"";
} else {
   echo "The regex pattern does not match. :(";
}

?>

edited Sep 19, 2016 at 17:52

answered Sep 19, 2016 at 16:27

A_Elric

3,59816 gold badges55 silver badges89 bronze badges

Sign up to request clarification or add additional context in comments.

12 Comments

jose Over a year ago

How about turning ![enter image description here][1] [1]: somewebsite.com/images/mycat.jpg into <img src="somewebsite.com/images/mycat.jpg ">

A_Elric Over a year ago

You want to extract the part after src?

A_Elric Over a year ago

Added to pull from src

jose Over a year ago

No actually your answer is what I needed, to extract the url part, what happens is that I have a dynamic text, example "This is text that is going to be changin and it could have a reference to an image like this ![enter image description here][1] [1]: somewebsite.com/images/mycat.jpg so i need to turn that reference into an actual image"

jose Over a year ago

Note that h-ttp:// part is erased by stackoverflow after the last [1]:

|

Collectives™ on Stack Overflow

replace a pattern to extract an image url from data php

1 Answer 1

12 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

12 Comments

Your Answer

Sign up or log in

Post as a guest

Related