CODEWITHSUNDEEP
phpjqueryjson
Hansen
Hansen

Reputation: 690

parsing json error : SyntaxError: JSON.parse: unexpected character at line 1 column 2 of the JSON data

i have a problem when parsing json from php to javascript

this is my example code :

//function
MethodAjax = function (wsFile, param) {
    return $.ajax({
        type: "POST",
        dataType: "json",
        url: '../proses/' + wsFile + ".proses.php",
        data: 'param='+param,
        error: function (msg) {
            return;
        },
    });
};

//call function 
$(document).ready(function() {

    $('#getproduk').click(function(){
        var param = {
        ProdukId : '1',
        ProdukName : 'test'
    };

    CallMethodWithAjax('try', JSON.stringify(param)).done(function(data){
        $data =  JSON && JSON.parse(data) || $.parseJSON(data); 
    });
});

//Simple Php code
<?php
    $data = $_POST['param'];

    $data = (json_decode($data));

    $data1['name'] = $data->ProdukName;
    $data1['id'] = $data->ProdukId;
    $data1['test'] = 'test';


    echo json_encode($data1);
?>

//post, response and error at console
response : {"name":"test","id":"1","test":"test"}
post : param    {"ProdukId":"1","ProdukName":"test"}
error : SyntaxError: JSON.parse: unexpected character at line 1 column 2 of the JSON  data

how to solve this probled, i have try the solution that i found at SO and google, but still cannot solve this problem

please someone help

thanks

Upvotes: 1

Views: 4871

Answers (1)

Bart Jedrocha
Bart Jedrocha

Reputation: 11580

jQuery's $.ajax() function will yield a JavaScript object if the response is JSON so I believe the error you're seeing is a result of trying to parse a JavaScript object and not a string as you're expecting. In the callback you're providing to the done function, inspect data and you'll find that it's an object and there is no need to JSON.parse the result.

Upvotes: 3

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