PHP JSON pass as query string

Question

I have few status return into json_encode object based on certain circumstances.

if(verify == 1)
    $data = array('status'=>'SUCCESS', 'points'=>$points, 'user'=>$user);

if(verify == 2)
    $data = array('status'=>'INACTIVE');

if(verify == 0)
    $data = array('status'=>'FAILED');

$data_str = json_encode($data);

I need $data_str to add as query string, when it redirect to hitter URL, such as: https://www.example.com/members?status=SUCCESS&points=2500&[email protected] OR https://www.example.com/members?status=INACTIVE

How could $data_str to be pass as query string?

You can use a foreach loop to loop through key-value pairs of $data and construct your query string. Don't forget to set rawurlencode for the value part. — Timothy Ha
– Timothy Ha, Commented May 16, 2016 at 9:51

Ali · Accepted Answer · 2016-05-16 09:59:11Z

3

You can use the PHP function http_build_query to achieve this and you never need to use anything else like foreach loop:

if($verify == 1)
    $data = array('status'=>'SUCCESS', 'points'=>$points, 'user'=>$user);

if($verify == 2)
    $data = array('status'=>'INACTIVE');

if($verify == 0)
    $data = array('status'=>'FAILED');

$url = https://www.example.com/members?.http_build_query($data);

EDIT

Here is demo

edited May 16, 2016 at 9:59

answered May 16, 2016 at 9:50

Ali

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2Fwebd · Accepted Answer · 2016-05-16 09:50:34Z

0

Another option would be :

Turn your array into a string using : implode
Make it ready for an URL using : urlencode

Hope that helps,

answered May 16, 2016 at 9:50

2Fwebd

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Comments

Murad Hasan · Accepted Answer · 2016-05-16 09:53:48Z

0

Try this:

$verify = 1;
$points = 2500;
$user = 'albert';

if($verify == 1)
    $data = array('status'=>'SUCCESS', 'points'=>$points, 'user'=>$user);

if($verify == 2)
    $data = array('status'=>'INACTIVE');

if($verify == 0)
    $data = array('status'=>'FAILED');

//$data_str = json_encode($data);

$qryStr = array();
foreach($data as  $key => $val){
    $qryStr[] = $key."=".$val;
}
echo $url = 'https://www.example.com/'.implode("&", $qryStr); //https://www.example.com/status=SUCCESS&points=2500&user=albert

OR use http_build_query().

echo $url = 'https://www.example.com/'.http_build_query($data); //https://www.example.com/status=SUCCESS&points=2500&user=albert

answered May 16, 2016 at 9:53

Murad Hasan

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