I have the following string:
"http://add.co" id="num_1"
How can I get only "http://add.co" from this string?
I tried to use pattern:
^http:\/\/(+s)*
4 Answers
You could use this regex ^"http:\/\/[^"]+(?=") which almost captures your url.
String : "http://add.co" id="num_1"
Matches : "http://add.co
You could append a last " to the match to fix it. Maybe someone can edit my regex to include the last ".
See example here: https://regex101.com/r/oppeaQ/1
Comments
There are a couple of ways to achieve what you want:
With preg_match:
$str = '"http://add.co" id="num_1"';
preg_match('/^"(.*?)"/', $str, $matches);
echo $matches[1];
With str_replace and explode:
$str = '"http://add.co" id="num_1"';
$url = str_replace("\"", "", explode(" ", $str)[0]);
echo $url;
Comments
<?php
$string = '"http://add.co id="num_1"';
preg_match_all('/[a-z]+:\/\/\S+/', $string, $matches);
print($matches[0][0]);
?>
o/p
http://add.co //tested in my machine
Comments
Like this:
1)The explode() function breaks your string into array based on seperator.
2).The preg_replace() replaces the contains of string matched with defined regular expression.
$string = '"http://add.co" id="num_1"';
$array = explode(':',$string);
echo preg_replace('/\"/','',$array[0]);
Output:
http
AND
$string = '"http://add.co" id="num_1"';
$array = explode(' ',$string);
echo preg_replace('/\"/','',$array[0]);
Output:
http://add.co
explode(" ", $str)[0]. Or can there be strings with no URL at the start? Then try'~^"?https?://\S+)*~'httporhttp://add,co.. See answer.