PHP: Stripping protocol from Instagram API image urls

Question

This is a snippet from an Instagram JSON:

"images":{
    "standard_resolution":
        { "url":"http:\/\/scontent-a.cdninstagram.com\/hphotos-xfa1\/t51.2885-15\/10593467_370803786404217_1595289732_n.jpg","width":640,"height":640}
}

I'd like to strip the protocol from ['images']['standard_resolution']['url']. I tried with:

.parse_url($value['images']['standard_resolution']['url'], PHP_URL_HOST) . parse_url($value['images']['standard_resolution']['url'], PHP_URL_PATH).

But does nothing! I think it has to do with the / escaping done in JSON (http:\/\/). Because if I try

.parse_url("http://www.google.com", PHP_URL_HOST) . parse_url("http://www.google.com", PHP_URL_PATH).

.. it just works fine. I'd like to keep things easy.. and don't use regex. parse_url would be perfect.

You could simply use [`str_replace("\/", "/", "$value['images']['standard_resolution']['url']")`](http://php.net/manual/en/function.str-replace.php)... OR, [`stripslashes()`](http://php.net/manual/en/function.stripslashes.php) — Enissay, Sep 27 '14 at 12:54

score 1 · Accepted Answer · answered Sep 27 '14 at 14:37

Why would you even need to do a replace or regex at all?

If you use json_decode the escaped slashes are un-escaped.

So doing like this:

<?php

$foo = '{"images":{"standard_resolution":{"url":"http:\/\/scontent-a.cdninstagram.com\/hphotos-xfa1\/t51.2885-15\/10593467_370803786404217_1595289732_n.jpg","width":640,"height":640}}}';
$bar = json_decode($foo, true);

$baz = 
parse_url($bar['images']['standard_resolution']['url'], PHP_URL_HOST) . 
parse_url($bar['images']['standard_resolution']['url'], PHP_URL_PATH);

echo $baz;

Would output:

scontent-a.cdninstagram.com/hphotos-xfa1/t51.2885-15/10593467_370803786404217_1595289732_n.jpg

See:

http://ideone.com/F0c4m9

PHP: Stripping protocol from Instagram API image urls

1 Answers1