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

Answer 1

您需要$_POST['data'] ，而不是$_POST['jsonParam'] 。

Answer 2

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);

Answer 3

{jsonParam: jsonParam,}代替{data: jsonParam,}

Answer 4

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

Answer 5

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.