Uncaught SyntaxError: Unexpected token < ajax call jsonp
Question
As the title states, getting this error in Chrome remote debugging. I am trying to send an ajax request (jsonp) to my .php file in localhost which will then do something to the database using the URL in the QR Code after a QR code is scanned. However, I am getting this error.
I am aware that jsonp is different from json and uses different syntax, however the code I am using worked for other ajax calls. I am unable to figure out the problem, and would appreciate some help.
Here are the codes:
.html file
<!DOCTYPE html>
<html>
<head>
<meta name="format-detection" content="telephone=no">
<meta name="msapplication-tap-highlight" content="no">
<meta name="viewport" content="user-scalable=no, initial-scale=1, maximum-scale=1, minimum-scale=1, width=device-width">
<title></title>
</head>
<body>
<div data-role="page" id="home">
<div data-role="header">
<h1></h1>
</div>
<div data-role="main" class="ui-content">
<p>
<a target="_blank" href="javascript:scan();" style="text-decoration: none"><button>Scan</button></a>
</p>
</div>
</div>
<div data-role="page" id="display">
<div data-role="header">
<h1>Display</h1>
</div>
<div data-role="main" class="ui-content">
<table data-role="table" data-mode="column" id="allTable" class="ui-responsive table-stroke">
<thead>
<tr>
<th>Name</th>
</tr>
</thead>
<tbody>
</tbody>
</table>
</div>
</div>
<script src="https://code.jquery.com/jquery-2.1.4.min.js"></script>
<script src="http://code.jquery.com/mobile/1.4.5/jquery.mobile-1.4.5.min.js"></script>
<script type= "text/javascript" src="js/jquery-3.1.1.js"></script>
<script type="text/javascript" src="cordova.js"></script>
<script>
function scan()
{
cordova.plugins.barcodeScanner.scan(
function (result) {
if(!result.cancelled)
{
if(result.format == "QR_CODE")
{
var value = result.text;
$.ajax({
type: "GET",
url: value + '?callback=?',
dataType: 'JSONP',
async: false,
jsonp : "callback",
jsonpCallback: "jsonpcallback",
success: function jsonpcallback(response)
{
if (response == "Success")
{
alert(response);
}
else
{
alert(response);
}
}
});
}
}
},
function (error) {
alert("Scanning failed: " + error);
}
);
}
</script>
</body>
</html>
.php file
<?php
header('Content-Type: application/json');
require 'dbcon.php';
session_start();
$acc_points = $_SESSION["acc_points"];
$acc_id = $_SESSION["acc_id"];
$result = $con->prepare(" UPDATE `points` SET `acc_points` = acc_points+1 WHERE `acc_id` = ? ");
$result->bind_param("i", $acc_id);
$result->execute();
if($acc_points != null)
{
$response = "Success";
echo $_GET['callback'] . '(' . json_encode($response) . ')';
}
else
{
$response = "Failed. Please try again.";
echo $_GET['callback'] . '(' . json_encode($response) . ')';
}
//connection closed
mysqli_close ($con);
?>
1 answers
solution1
0 2017-02-05 09:53:50
Error was:
: Uncaught exception 'mysqli_sql_exception' with message 'Duplicate entry '12' for key 'PRIMARY'' in C:\\xampp\\htdocs\\MP\\appqrcode.php:14 Stack trace: #0 C:\\xampp\\htdocs\\MP\\appqrcode.php(14): mysqli_stmt->execute() #1 {main} thrown in on line ：未捕获异常'mysqli_sql_exception'，消息'重复条目'12'用于C：\\ xampp \\ htdocs \\ MP \\ appqrcode.php中的键'PRIMARY'：14堆栈跟踪：＃0 C：\\ xampp \\ htdocs \\ MP \\ appqrcode.php（14）：在第行的抛出mysqli_stmt-> execute（）＃1 {main}
Solved the problem by changing the primary key of the table.
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