PHP: Getting associative array returns null, despite existing in var_dump

Question

I have an array of 'servers' that I'm storing in a JSON file.

The JSON file looks like this: {"1":{"available":1,"players":0,"maxplayers":4}}

I retrieve this value with this:

$servers = (array) json_decode(file_get_contents("activeservers.json"));

However, when I try to access the array with $server = $servers[$id], $server is null. I noticed that the key is in quotes, so I also tried casting $id to a string, and surrounding it with quotes (") which didn't work.

Something to note, is that this code is returning "NULL":

foreach(array_keys($servers) as $key){
    var_dump($servers[$key]);
}

$servers[$id] doesn't exist in you json. Where is the 'id' key in the json? — bos570
– bos570, Commented Jan 11, 2017 at 15:54
Instead of type casting just pass TRUE as second parameter to json_decode() — Rizier123
– Rizier123, Commented Jan 11, 2017 at 15:55
Yes, you're using it as a key but where are you setting value for $id? — bos570
– bos570, Commented Jan 11, 2017 at 15:59

André Ferraz · Accepted Answer · 2017-01-11 16:05:15Z

2

Your code is wrong. Also you don't need to type cast when doing a json_decode, you can instead set the second parameter to true more info here. Also you don't need to use the array_keys function in your foreach loop, try this.

$json    = '{"1":{"available":1,"players":0,"maxplayers":4}}';
$servers = json_decode($json, true);

foreach($servers as $key => $value) {
    print $value["available"];
}

Do a print_r($value) to get all the array keys available to use. Also you could take advantage of the $key variable to print out the array key of the parent array.

edited Jan 11, 2017 at 16:05

answered Jan 11, 2017 at 16:00

André Ferraz

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7 Comments

NBTX Over a year ago

Why print over echo?

André Ferraz Over a year ago

Its just a matter of preference. The differences are small: echo has no return value while print has a return value of 1 so it can be used in expressions.

NBTX Over a year ago

FYI: The foreach loop was just to demonstrate my point and wouldn't be used in production.

André Ferraz Over a year ago

@SamMearns it doesn't change the fact your code was incorrect.

André Ferraz Over a year ago

Yes I understand it but it is still pointless using it. Because the var_dump($servers[$key]); you're referencing is not the array_keys($servers) in your foreach loop. To prove my point is you do this foreach($test_servers = array_keys($servers) as $value) { print $test_servers[$value]; } your code will break. Am on this because I don't think you understood the use of array_keys, otherwise you would not use it even if it was for demonstration purposes. Am not picking on you by the way.

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NBTX · Accepted Answer · 2017-01-11 16:00:24Z

1

Thanks, @Rizier123 (who solved the question).

Apparently passing TRUE as the second parameter to my json_decode function fixes the issue.

After checking the PHP documentation for json_decode() (PHP: json_decode), it seems that passing this parameter means that the resulting decoded array is automatically converted into an associative array (and this is recurring, meaning that this automatically happens for sub-arrays).

Edit: @Rizier123 also says that "you might want to read: stackoverflow.com/a/10333200 to understand a bit better why it is so "weird" and your method didn't work properly."

answered Jan 11, 2017 at 16:00

NBTX

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