CODEWITHSUNDEEP
phparraysencodingjson
Mihai Bujanca
Mihai Bujanca

Reputation: 4219

json_decode to an array?

foreach ($likes as $like) {
    // Extract the pieces of info we need from the requests above
    $id = idx($like, 'id');
    $item = idx($like, 'name');

    fwrite($fileout,json_encode($like));
    fwrite($fileout,PHP_EOL );
}

$json_string = file_get_contents('testson.json');
$get_json_values=json_decode($json_string,true);

foreach ($get_json_values as $getlikes) {  ?>
          <li>
            <a href="https://www.facebook.com/<?php echo $getlikes['id']; ?>" target="_top">
          </li>
        <?php
      }

When opening the browser, there is a Warning: Invalid argument supplied for foreach(). I don't understand why would my arguments be invalid.

If I add the if, nothing happens, which shows what the actual problem is. But the question is WHAT IS THE PROPER WAY TO DO THIS? I'm pretty sure it's very simple, but i've been struggling with this for more than an hour. My json files has fields like this, so I don't think there would be the problem:

{"category":"Musician\/band","name":"Yann Tiersen (official)","id":"18359161762"}

Please help me, I really got tired with it, and I don't know what to do. So... how can I decode the file into an array?

Upvotes: 1

Views: 344

Answers (2)

powtac
powtac

Reputation: 41050

You need the contents of the testson.json file not just the name!

Receive the contents via PHP:

$json_string = file_get_contents('testson.json');

Make sure there are valid JSON contents in the file by testing it via

var_dump(json_decode($json_string));

And then call 

foreach (json_decode($json_string) as $getlikes) {  ?>

Update: When you save the file and access it miliseconds later it might be that the filesystem is too slow and not showing you the correct content.

Add

fclose($fileout);
clearstatcache();

before the file_get_contents(); call!

I recommend to use file_put_contents() and file_read_contents() to get rid of such problems!

Upvotes: 2

Bafsky
Bafsky

Reputation: 771

json_decode is expecting a string as it's first parameter. You are passing what I'm guessing is a filename. You should load the file first like so...

$json = file_get_contents('testson.json');
$data = json_decode($json);

Upvotes: 1

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