How do I parse a JSON object in PHP?

Question

I've got a JSON object that I'm sending to a PHP script and I'm having trouble parsing the JSON. Here's the POST request:

http://mywebsite.com?action=somefunction&{%22id%22:1,%22Name%22:%22Mike%22}

And here's my PHP function, which obviously doesn't work:

$data = $_GET['data'];
$obj = json_decode($data);
echo $obj->Name;
die();

The end goal is to extract the name "Mike" from the URL string. Any suggestions?

Do a var_dump on $obj.

Daniel
– Daniel

2013-04-14 00:27:39 +00:00
Commented Apr 14, 2013 at 0:27 — Daniel
– Daniel, Commented Apr 14, 2013 at 0:27
Your query string is wrong.. there is no data=.

Esailija
– Esailija

2013-04-14 00:29:11 +00:00
Commented Apr 14, 2013 at 0:29 — Esailija
– Esailija, Commented Apr 14, 2013 at 0:29
You're right. What should this be instead?

kaleazy
– kaleazy

2013-04-14 00:33:29 +00:00
Commented Apr 14, 2013 at 0:33 — kaleazy
– kaleazy, Commented Apr 14, 2013 at 0:33

Amber · Accepted Answer · 2013-04-14 00:27:32Z

5

Try taking a look at what PHP is outputting from json_decode():

$data = $_GET['data'];
$obj = json_decode($data);
var_dump($obj);

Your code itself works fine: http://ideone.com/0jsjgT

But your query string is missing the data= before the actual JSON. This:

http://mywebsite.com?action=somefunction&{%22id%22:1,%22Name%22:%22Mike%22}

should be this:

http://mywebsite.com?action=somefunction&data={%22id%22:1,%22Name%22:%22Mike%22}

answered Apr 14, 2013 at 0:27

Amber

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2 Comments

kaleazy Over a year ago

Ahh, you're right! But even with data set in the URL, why doesn't this work? $data = $_GET['data']; $obj = json_decode($data);

kaleazy Over a year ago

Thanks for the help guys. Turned out I needed stripslashes($data), then it worked!

Community · Accepted Answer · 2017-05-23 12:06:07Z

1

you should do

echo $obj->{'Name'};

This also is a duplicate question of Echo data json by json_decode

edited May 23, 2017 at 12:06

CommunityBot

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answered Apr 14, 2013 at 0:27

KennyV

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3 Comments

hek2mgl Over a year ago

What is the difference to call it without {} ??

Amber Over a year ago

$obj->Name and $obj->{'Name'} are equivalent in this case; this answer is useless.

Esailija Over a year ago

@hek2mgl there is none (in this case) :P

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