PHP syntax: same snippets in same file return different values

Question

I'm having a hard time with PHP formatting. :P In a function I have these strings and $cache1 path is output properly (wordpress absolute upload path + /json/dailymile/BLABLA.json

$url1 = "http://api.dailymile.com/people/XXXXX/entries.json";
$cache1 = $upload_dir['basedir'].'/json/dailymile/'.sha1($url1).'.json';

Then I have a SIMILAR (if not the same) case but only God knows why, $cache2 returns just /json/instagram/BLABLA.json without wordpress upload path before.

$url2 = 'https://api.instagram.com/v1/users/'.$user_id.'/media/recent/?access_token='.$token.'&count='.$count;
$cache2 = $upload_dir['basedir'].'/json/instagram/'.sha1($url2).'.json';

These two snippets are in the same file and the top of this file I have <?php $upload_dir = wp_upload_dir(); ?>.

Why the second snippet doesn't return the WP upload dir before /json/instagram/BLABLA.json ?

Try inserting var_dump($upload_dir); right before the second snippet starts. — ComFreek
– ComFreek, Commented Dec 29, 2013 at 15:05
@ninetwozero added the whole function. ComFreek: ok, trying right now — MultiformeIngegno
– MultiformeIngegno, Commented Dec 29, 2013 at 15:12
@ComFreek This is the output: pastebin.com/raw.php?i=9PmCQq9d — MultiformeIngegno
– MultiformeIngegno, Commented Dec 29, 2013 at 15:17

akirk · Accepted Answer · 2013-12-29 15:15:33Z

1

$upload_dir is outside of the scope when using your code within a function. Using the global keyword you can import the variable from the global scope into the current one. Like this:

<?php
function get_instagram(...) {
    global $upload_dir;
    ...
}

answered Dec 29, 2013 at 15:15

akirk

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4 Comments

ComFreek Over a year ago

While this works, one should rather pass $upload_dir as a parameter.

akirk Over a year ago

It depends on the use case and the quality of code you want to achieve. In this case it looks like a "quick hack", we have a function with plenty parameters already. You would probably first want to get rid of all the parameters by wrapping it in a class before you add yet another parameter.

MultiformeIngegno Over a year ago

But why I have other functions in the same file and instead for them it works also if it's not inside the function..?

akirk Over a year ago

Maybe in the other functions $upload_dir is defined before being used?

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