Question

我不知道我做错了什么。我有一个对话框，用户可以输入地址并单击确定，然后将输入的信息插入到数据库中。我不断收到“意外的输入结束”错误或“意外令牌”错误。我觉得它与jquery ajax函数中的serialize属性有关，但我不确定。

我不知道如何回显变量以查看它们是否甚至从字段或sql语句中获取数据。我已经尝试了很多，它或者崩溃了页面，或者它给了我同样的“意外”错误。

请帮忙！：（

jQuery代码：

    $('#ChangeOfAddress').click(function() {
        //change of address dialog
        $( "#ChangeAddressDialog" ).dialog({
            width:500,
            modal:true,
            closeOnEscape:true,
            buttons: [ 
                { text: "Ok", type: "submit", click: function() { 
                        $.ajax({
                            url: "classes/add-address.php",
                            type: "POST",
                            data: $("#main_form").serialize(),
                            dataType: 'json',
                            error: function(SMLHttpRequest, textStatus, errorThrown){
                                alert("An error has occurred making the request: " + errorThrown)

                            },
                            success: function(result){
                                //do stuff here on success such as modal info
                                //$("#main_form").submit();
                                $(this).dialog("close");
                            }
                        })
                    } 
                },
                { text: "Close", click: function() { $(this).dialog( "close" ); } } ]
        });//end dialog
    });

PHP页面:( ez sql insert返回bool值）

<?php
require_once('../config.php');

$sqlCheck = ''; 
$parcel_id = isset($_POST['ParcelId']) ? $_POST['ParcelId'] : null;
$address1 = isset($_POST['Address1']) ? $_POST['Address1'] : null;
$address2 = isset($_POST['Address2']) ? $_POST['Address2'] : null;
$city = isset($_POST['City']) ? $_POST['City'] : null;
$state = isset($_POST['State']) ? $_POST['State'] : null;
$zip = isset($_POST['Zip']) ? $_POST['Zip'] : null;
$country = isset($_POST['Country']) ? $_POST['Country'] : null;

$db = new ezSQL_mysql(DB_USER, DB_PASSWORD, DB_NAME, DB_HOST);
$result = $db->query("INSERT INTO change_of_address (parcel_id, address_1, address_2, City, State, Zip, Country) VALUES ('" . $parcel_id . "','" . $address1 . "','" . $address2 . "','" . $city . "','" . $state . "','" . $zip . "','" . $country . "')");
if ($result == 1) {
    echo true;
} else {
    echo false;
}

//$sqlCheck = "INSERT INTO change_of_address (parcel_id, address_1, address_2, City, State, Zip, Country) VALUES ('" . $parcel_id . "','" . $address1 . "','" . $address2 . "','" . $city . "','" . $state . "','" . $zip . "','" . $country . "')";



?>

我尝试添加以下内容，但它没有返回任何错误：

error_reporting(E_ALL);
ini_set('display_errors', 'On');

Answer 1

您的问题是您的$.ajax调用期待JSON响应，但您的PHP文件正在回显文字true或false。这些不是JSON响应。您需要返回有效的JSON，例如

if ($result == 1) {
    echo '{"success":true}';
} else {
    echo '{"success":false}';
}

意外结束输入错误，再次

1 个答案: