PHP and JSON - Decoded Array Issue?

Question

I'm trying to get some info out of a decoded JSon string into an array.

I've this code:

$json_d='{ // This is just an example, I normally get this from a request...
             "iklive.com":{"status":"regthroughothers","classkey":"domcno"}
}';

$json_a=json_decode($json_d,true);
$full_domain = $domain.$tlds; // $domain = 'iklive' ; $tlds = '.com'

echo $json_a[$full_domain][status];

The problem is, I need to get the value for "status" of "iklive.com" but when I do echo $json_a[$full_domain][status]; it does not work, but if I do it manually like echo $json_a['iklive.com'][status]; (with the quotes there) it works.

I've tried to add the quotes to the variable but without success, how can I do this?

Thanks Everyone!

Thanks to Pekka and jeromegamez I noticed a error in the HTML part of this "problem", the $tlds variable was "com" instead of ".com" -- Sorry by wasting your time with this. I do feel bad now.

Anyway, thanks to jeromegamez and Marc B I discovered that unless status was a constant I need to quote it ;) You can check jeromegamez answer to a detailed explanation of the problem and proper debug.

Sorry.

You are 100% positive $full_domain has that value? Then there shouldn't be a problem. Are you keeping in mind that array indexes are case sensitive? — Pekka
– Pekka, Commented Nov 3, 2011 at 21:04
Actually, neither should work, because status hasn't been quoted. Unless you've got a status constant define()'d, a PHP install in strict mode should gripe about both versions. — Marc B
– Marc B, Commented Nov 3, 2011 at 21:04
What do you get exactly when you echo $full_domain ?? And also make sure there's no spaces in there. — aziz punjani
– aziz punjani, Commented Nov 3, 2011 at 21:09

jeromegamez · Accepted Answer · 2011-11-03 21:30:02Z

2

This works for me:

<?php
$json_d='{ "iklive.com":{"status":"regthroughothers","classkey":"domcno"} }';
$json_a = json_decode($json_d, true);

if (!is_array($json_a)) {
    echo "\$json_d is not a valid JSON array\n";
}

$domain = "iklive";
$tld = ".com";
$full_domain = $domain . $tld;

if (!isset($json_a[$full_domain])) {
    echo "{$full_domain} is not set in \$json_a\n";
} else {
    echo $json_a[$full_domain]['status']."\n";
}

What I did:

Changed json_a[$full_domain][status] to json_a[$full_domain]['status'] - the missing quotes around status don't break your script, but raise a Notice: Use of undefined constant status - assumed 'status'
Added a check if the decoded JSON is actually an array
Added a check whether the key $full_domain is set in $json_a

edited Nov 3, 2011 at 21:30

answered Nov 3, 2011 at 21:22

jeromegamez

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2 Comments

TCB13 Over a year ago

You're the best! I discovered by problem was a stupid mistake done in the HTML file that posts the $domain and $tld variables. Sorry by wasting your time guys. Anyway, I noticed also I was not using ['status'] quoted! Thanks! ;)

pal4life Over a year ago

Helped me for Drupal Json Decode as well where it was giving a bunch of similar error messages

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