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I have following code in php:
$link ='mylink';
$description = '<p><img src="http://l3.yimg.com/bt/api/res/1.2/hmYETKsJS2CXBsG7oO125w--/YXBwaWQ9eW5ld3M7Zmk9ZmlsbDtoPTg2O3E9ODU7dz0xMzA-/http://media.zenfs.com/en_us/News/Reuters/2013-09-07T045234Z_3_CBRE9851BPL00_RTROPTP_2_RUSSIA-G20.JPG">By Roberta Rampton and Susan Cornwell WASHINGTON (Reuters) - ...</p>';
$pattern = "/(?<=href=(\"|'))[^\"']+(?=(\"|'))/";
$new_des = preg_replace($pattern,"?url=".$link,$description);
How can I do that in javascript?
pattern = /(href=["'])[^"']+(?=["'])/
new_des = description.replace(pattern, "$1?url=" + link)
JS doesn't support lookbehinds, therefore you'll have to capture the first group and insert it back when replacing using $1.
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i have this string
$string = "Social\Notify\Models\User";
How can i tell to php how select just the fourth segment of it? in this case just the word User?
Something like this will do?
$str='Social\Notify\Models\User';
echo explode('\\',$str)[3]; //"prints" User
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How to match the following link structure that started with rtmp://
$link = "rtmp://00.000.000.000/live"; // GOOD
$link = "http://00.000.000.000/live"; // BAD
$link = "www.00.000.000.000/live"; // BAD
so only matching the links with rtmp with preg_match or whatever ~ thanks
I'd do:
preg_match('~^rtmp://~', $link);
Something like this will suffice..
if(stripos($link,'rtmp://')!==false)
{
echo "Good";
}
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php script query out put is
"some word STAR some other words", I want remove all workd before "Star". I need all other words after "Star".
I using php script output from a mysql file
$newStr = substr_replace($str, '', 0, strpos($str,"STAR"));
Where $str is a variable with a string, eg: "some word STAR some other words"
$arrayStr = explode('STAR', 'some word STAR some other words');
echo $arrayStr[1];
DEMO
Some documentation:
php.net explode() function
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I need to create a regex that will select the data between two characters (& - #).
& foo #
I would like to select foo.
Any ideas? Good explanations are also appreciated.
how about this:
preg_match('~&([^#]*)#~',$content,$match);
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I have text like this Too early to post? http://www.somewebsite.com/article/226973 I want to parse the hyperlinks in the text and make the text look like so Too early to post? http://www.entrepreneur.com/article/226973 I want to do this but I have no idea where to start from or what regexp to use.
$s = 'Too early to post? https://www.somewebsite.com/article/226973';
$parsed = preg_replace('#(https?://([-\w\.]+[-\w])+(:\d+)?(/([\w/_\.#-]*(\?\S+)?[^\.\s])?)?)#', '$1', $s);
echo $parsed;