PHP json_encode failing

Question

I'm trying to read this in Objective-C but I keep getting errors.

When I run the json through a validator it tells me it is ok. But json_decode gets a null value.

What am I missing?

$arrTest = array("key" => "This is a string");
echo json_encode($arrTest);

$ob =  json_decode($arrTest);

if ($ob === NULL) { 
    print "\nDang it";
}

$arrTest is an array, not a string. You can't decode an array. — gen_Eric
– gen_Eric, Commented Apr 7, 2014 at 15:46

p.s.w.g · Accepted Answer · 2014-04-07 15:46:40Z

6

json_encode returns a new string, so you have to save that to a variable if you want to decode it later:

$arrTest = array("key" => "This is a string");
$jsonString = json_encode($arrTest);
echo $jsonString;

$ob =  json_decode($jsonString);

if ($ob === NULL) { 
    print "\nDang it";
}

answered Apr 7, 2014 at 15:46

p.s.w.g

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1 Comment

PruitIgoe Over a year ago

Thanks everyone, it's been a long time since I've done any web front end work.

lukasrozs · Accepted Answer · 2014-04-07 15:47:39Z

1

You're trying to decode the array, not the json string

answered Apr 7, 2014 at 15:47

lukasrozs

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Comments

gbestard · Accepted Answer · 2014-04-07 15:48:21Z

1

You are decoding the array not the encoded string.

answered Apr 7, 2014 at 15:48

gbestard

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