Question

i have a code to get a list of pattern from a website: https://noembed.com/providers

with the following code, i can display all the patterns:

$supported = 'https://noembed.com/providers' ;
$jsonSUPP = json_decode(file_get_contents($supported),true) ;
echo 'pour:'.$url.'<br/>'  ;
foreach ($jsonSUPP as $key => $value) { 
foreach ($value[patterns] as $ent) { 
echo "**patt : $ent <br />\n" ;
}
}

now i would like to create a condition: if $url is matching with $ent then …
i tried with preg_match but i have an error. and ereg too

could you help me please

Answer 1

Try and use preg_quote so to escape slashes in the url:

if(preg_match('/' . preg_quote($url) . '/', $ent))
   return true;

Or change delimiter:

if(preg_match('%' . $url . '%', $ent))
   return true