javascript - 无法从 PHP 数组中检索数据，该数组由使用 JQuery Ajax 的 function() 返回

Question

我有一个“点击监听器”，它调用并向函数“getModalData”发送一个 userId 参数，然后该函数将一个数组值返回到变量“arrayedUserData”。

$('body').on('click', '.openModal', function () {
        var userId = $(this).val(),
            btnText = $(this).text(),
            btnClass = '',
            colorCode = '',
            arrayedUserData = getModalData(userId);

        if (btnText === "Delete") {
            btnClass = 'danger';
            colorCode = '#d9534f';
        } else {
            btnClass = 'warning';
            colorCode = '#f0ad4e';
        }

        $('#actionBtn').removeClass().addClass('btn btn-' + btnClass).text(btnText);
        $('#modalTitle').text('Confirm ' + btnText);
        $('#S-modalbody p').text('Are you sure you want to ' + btnText + ' user: ');
        $('#S-modalbody h4').css('color', colorCode).text(userId + " - " + arrayedUserData.LastName + ", " + arrayedUserData.FirstName);

    });

这是函数“getModalData”。 Ajax“成功”返回的 php 数组将被传递给函数返回的变量“UserData”。

function getModalData(passedUserId) {
        var UserData;
        $.ajax(
            {
                type: "POST",
                url: "get/get_modal_data.php",
                data: { passedUserId: passedUserId },
                dataType: "json",
                success: function (data) {
                   UserData = data;
                }
            }
        );
        return UserData;
    }

这是“get_modal_data.php”。

<?php
    include "../includes/connect.php";

    if (isset($_POST['passedUserId'])) {
        $UserId = mysqli_real_escape_string($con, $_POST['passedUserId']);

        $getUserData = mysqli_query($con, "SELECT * FROM tblUserAccounts WHERE uaUserId = '".$UserId."'");
        $uaRow = mysqli_fetch_assoc($getUserData);

        $UserDataArr = array("UserId" => $uaRow['uaUserId'],
                     "EmailAddress" => $uaRow['uaEmailAddress'],
                     "FirstName" => $uaRow['uaFirstName'],
                     "LastName" => $uaRow['uaLastName'],
                     "BirthDate" => $uaRow['uaBirthDate'],
                     "Address" => $uaRow['uaAddress'],
                     "Gender" => $uaRow['uaGender'],
                     "ContactNumber" => $uaRow['uaContactNumber'],
                     "BloodTypeId" => $uaRow['uaBloodTypeId'],
                     "AccountStatus" => $uaRow['uaAccountStatus'],
                    );


        echo json_encode($UserDataArr);
        exit();
    }
?>

控制台上出现此错误:

Uncaught TypeError: Cannot read property 'LastName' of undefined get_user_accounts.js:66

这是 get_user_accounts.js 的第 66 行，它出现在“点击监听器”上。

$('#S-modalbody h4').css('color', colorCode).text(userId + " - " + arrayedUserData.LastName + ", " + arrayedUserData.FirstName);

但是，我很困惑，因为 php 数组出现在浏览器的网络响应上:

Successful Connection{"UserId":"1","EmailAddress":"paulanselmendoza@gmail.com","FirstName":"Paul Ansel","LastName":"Mendoza","BirthDate":"1998-12-17","Address":"Phase 1B Block 8 Lot 20 Olivarez Homes South, Sto. Tomas, Binan City, Laguna","Gender":"Male","ContactNumber":"2147483647","BloodTypeId":"0","AccountStatus":"ACTIVE"}

Answer 1

您是否看到在 JSON 数据之前收到:Successful Connection ？您必须删除它，否则它将是无效的 JSON 响应。您共享的代码没有特定的内容。

我相信您必须检查您的数据库连接，在成功连接时，它被设置为输出成功连接，这会破坏您的响应。请删除那段代码。

include "../includes/connect.php";

它可以是这样的:

$conn = mysqli_connect() or die("Error");
echo "Successful Connection";