Is this json format?

Question

I am getting this output:

{"params":{"2":true,"3":true}}

I have printed this as (at PHP side):

$data = file_get_contents("php://input");
        print($data);

I want to use that 2 and 3 values using loop at PHP side. But as I am trying, it is not json format (its doubt). So how can i use these values using loop?

Looks like valid json, show the code you have that is failing, eg the loop — Steve, Jun 17 '15 at 14:57
http://stackoverflow.com/questions/4731242/php-loop-through-json-array Google search failure — FirebladeDan, Jun 17 '15 at 14:59
json is just a string in php, to use it as a data structure, you must convert it to an array or object with `json_decode()` — Steve, Jun 17 '15 at 15:00
Did you notice, here square brackets are not present, still it is json? I don't think so! If it is not json, how can I decode it with json_decode ? — VBMali, Jun 17 '15 at 15:02

score 3 · Accepted Answer · answered Jun 17 '15 at 15:01

3

$j='{"params":{"2":true,"3":true}}';

$decoded=json_decode($j,1);

print "<pre>\n";
print_r($decoded);
print "</pre>\n";

The second parameter of json_decode() is

'assoc' - When TRUE, returned object s will be converted into associative array

answered Jun 17 '15 at 15:01

KevInSol

2,560
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46

score 2 · Answer 2 · answered Jun 17 '15 at 15:11

2

<?php

$json='{"params":{"2":true,"3":true}}';

$params=json_decode($json, true)['params'];

foreach ($params as $k =>$v){
    echo $k . ' is ' .  var_export($v, true) . PHP_EOL;
}

output:

2 is true
3 is true

answered Jun 17 '15 at 15:11

Ali

2,993
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19
42

score 0 · Answer 3 · answered Jun 17 '15 at 15:15

$data='{"params":{"2":true,"3":true}}';
$decoded=json_decode($data,1);
        foreach ($decoded as $key => $param)
        {
            foreach ($param as $user_id => $value)
            {
                    echo $user_id;
            }
        }

It gives me output:

2 3

(as I expected).

Is this json format?

3 Answers3