Decode JSON with PHP

Question

Can some one tell why this simple JSON with JqUERY doesn't work for me?

I have this JS code,

var jsonParam = <? $json = $_SESSION['searchSess']; echo json_encode($json);?>;
jsonParam = JSON.stringify(jsonParam);
$(document).ready(function(){
      $.post("searching.php?rdr=search", {data: jsonParam,}, function (data){
        alert(data)
    })
});

And Here is the PHP code,

$data = json_decode($_POST['jsonParam'], true);
var_dump($data);

And the response is null or nothing,

can please someone help whats wrong here?

Thank you

Niet the Dark Absol · Accepted Answer · 2012-08-23 01:21:05Z

4

You want $_POST['data'], not $_POST['jsonParam'].

answered Aug 23, 2012 at 1:21

Niet the Dark Absol

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Comments

Michael Berkowski · Accepted Answer · 2012-08-23 01:20:53Z

3

jsonParam was the JavaScript variable, but it was posted to PHP as $_POST['data'] since you passed {data: jsonParam} into the $.post.

// Instead:
$data = json_decode($_POST['data'], TRUE);
var_dump($data);

answered Aug 23, 2012 at 1:20

Michael Berkowski

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6 Comments

thegrede Over a year ago

Thank you, but this also returns null, :(

Michael Berkowski Over a year ago

var_dump($_POST) to see if you got anything, and also look at your page source to be sure that the original json_encode($json) printed what it was supposed to.

thegrede Over a year ago

Ok, with var_dump($_POST) I'm getting this [data] => {\"locations\":\"ARP\"}, so how do I perse the array?

Michael Berkowski Over a year ago

@thegrede json_decode($_POST['data'], TRUE) handles that just fine for me, $data["locations"] is ARP

thegrede Over a year ago

But when I do json_decode($_POST['data'], TRUE) I'm getting nothing, :(

|

user849137 · Accepted Answer · 2012-08-23 01:23:11Z

2

{jsonParam: jsonParam,} Instead of {data: jsonParam,}

answered Aug 23, 2012 at 1:23

user849137

Comments

Mahn · Accepted Answer · 2012-08-23 02:07:07Z

1

Try the following:

JS:

var jsonParam = <? 
    $json = $_SESSION['searchSess']; 
    $json['longitude'] = (string) $json['longitude'];
    $json['latitude'] = (string) $json['latitude'];

    echo json_encode($json);
?>

$(document).ready(function(){
      $.post("searching.php?rdr=search", {data: jsonParam }, function (data){
        alert(data)
    })
});

PHP:

$data = json_decode($_POST['data'], true);
var_dump($data);

I suspect your longitude and latitude fields are not being parsed correctly as floats.

edited Aug 23, 2012 at 2:07

answered Aug 23, 2012 at 1:37

Mahn

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8 Comments

thegrede Over a year ago

Nops!! :( it retruns empty too, I dont know why, i'm copying it and paste too and no string returns, brrrrr.

Mahn Over a year ago

Updated, try again, I missed the quotes on JSON.parse.

thegrede Over a year ago

Ye, I tried that with the quotes before too, but the same :( no stritng returns, I guess its somethings else where, and need to be figure out, does it works when psste it in your server?

Mahn Over a year ago

I don't know what your $_SESSION['searchSess'] is so it's hard for me to tell if it would work :) can you post what you get from doing echo json_encode($json); on a blank page?

thegrede Over a year ago

This is the result from a blank page,

{"locations":"ARP","pickupair":"JFK","dropoffair":"JFK","pickUpLocation":"John F. Kennedy International Airport (JFK), 144-02 135th Ave, Queens, NY 11436, USA","dropOffLocation":"","pickupmonth":"08","drol":"false","pickupday":"23","pickupyear":"2012","dropoffmonth":"08","dropoffday":"30","dropoffyear":"2012","pickupmin":"09","pickuphour":"00","dropoffpmin":"09","dropoffhour":"00","latitude":4064384.36,"longitude":-7378230.35,"relatitude":0,"relongitude":0,"hourPU":"18","hourDO":"18"}

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Lusitanian · Accepted Answer · 2012-08-23 01:21:57Z

1

$data = json_decode($_POST['jsonParam'], true); should be $data = json_decode($_POST['data'], true);

answered Aug 23, 2012 at 1:21

Lusitanian

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