Question

这是我的jQuery ......

$.ajax({
                type: 'POST',
                url: 'http://adamscarter.co.uk/daily/facebook/savefbprofile.php',
                data: {'fb_id': fb_id, 'fbusername': fbusername, 'location_id': location_id, 'gender': gender},
                beforeSend: function() {
                    console.log('Before send: ' + fb_id, fbusername, location_id, gender);
                },
                success: function(data) {
                    console.log('saveFbProfile post ajax success');
                    console.log(data);
                }
            });

...这是我的PHP ......

<?php
session_start();

include('dbcon.php');

$gender = substr($_POST['gender'], 0, 1);
$fb_id = $_POST['fb_id'];
$location_id = $_POST['location_id'];
$fbusername = $_POST['fbusername'];

//Set session vars
$_SESSION['gender'] = $gender;
$_SESSION['fb_id'] = $fb_id;
$_SESSION['location_id'] = $location_id;
$_SESSION['fbusername'] = $fbusername;

if (isset($_SESSION['userid'])) {
    mysql_select_db('users', $GLOBALS['conInsert']);
    $sql = "UPDATE user_fb_details SET gender = '". $gender. "', fb_id = '". $fb_id. "', location_id = '". $location_id. "', username = '". $fbusername. "' WHERE user_id = '". $_SESSION['userid']. "'";
    mysql_query($sql, $GLOBALS['conInsert']);
}
?>

我的代码中有错误吗？当我记录变量时，它们都具有正确的值，但我只是得到一个'parseerror'。

Answer 1

听起来jQuery可能会尝试将POST的结果解析为JSON响应而失败，因为它不是JSON。您可能希望将dataType设置为'text'，或者从PHP返回不同的Content-Type。

您肯定希望通过一些mysql_real_escape_string或参数化查询修复危险的SQL注入漏洞。

PHP Ajax jQuery Post - 我做错了什么？

1 个答案: