返回AJAX成功和錯誤

Question

我已經構建了一個使用AJAX提交表單數據的登錄腳本。

沒有AJAX，PHP部分工作正常。 但該系統不適用於AJAX實現。

它始終顯示以下消息，即使PHP文件返回true [正確的用戶名和密碼] ...似乎Jquery中的if條件不起作用。

用戶名/密碼不正確

HTML結果分組

<div id="user-result" align="center"></div>

jQuery的

<script type="text/javascript">
    $(document).ready(function () {
        var form = $('#loginform');
        form.submit(function (ev) {
            ev.preventDefault();
            $.ajax({
                type: form.attr('method'),
                url: form.attr('action'),
                cache: false,
                data: form.serialize(),
                success: function (data) {
                    if (data == "true") {
                        $("#user-result").html("<font color ='#006600'> Logged in | Redirecting..</font>").show("fast");
                        setTimeout(
                            function () {
                                window.location.replace("index.php");
                            }, 1500);
                    } else {
                        $("#user-result").html("<font color ='red'> Incorrect Username/Password</font>").show("fast");
                    }

                }

            });
        });
    }); 
</script>

fn_login.php

<?php
{
    session_start();
    include_once 'db_connect.php';

    if (isset($_POST))
     {
        $email = filter_input(INPUT_POST, 'email', FILTER_SANITIZE_STRING);
        $logpwd = filter_input(INPUT_POST, 'password', FILTER_SANITIZE_STRING);

        $stmt = $conn->prepare("SELECT password FROM manager WHERE email = ? LIMIT 1");
        $stmt->bind_param("s", $email);


        $stmt->execute();
        $stmt->store_result();

        // get variables from result.
        $stmt->bind_result($password);
        $stmt->fetch();


        // Check if a user has provided the correct password by comparing what they typed with our hash
        if (password_verify($logpwd, $password))
        {
            $sql = "SELECT * from manager WHERE email LIKE '{$email}' LIMIT 1";
            $result = $conn->query($sql);
            $row=mysqli_fetch_array($result);
            $id = $row['id'];
            $conn->query("UPDATE manager SET lastlogin = NOW() WHERE id = $id");

            $_SESSION['manager_check'] = 1;
            $_SESSION['email'] = $row['email'];
            $_SESSION['fullname'] = $row['fullname'];
            $_SESSION['designation'] = $row['designation'];
            $_SESSION['id'] = $row['id'];
            echo "true";
        }
         else {
            die();
        }
     }
}

?>      

有人可以指出代碼/實踐中的錯誤。

EDIT

Just Tried禁用AJAX，當用戶名/傳遞正確時，PHP文件正常回顯true

Answer 1

你有空格?>

因此，AJAX響應在true之后有空格。

解：

從PHP文件的末尾刪除?> 。

它不會影響任何PHP功能。

而你的AJAX響應將沒有空格。

從PHP文件末尾排除結束標記?>是現代PHP框架和CMS的標准做法。

調試AJAX的提示：

1）始終在或Chrome上使用Firefox（使用Firebug Add）。

2）使用Firebug的Console選項卡，檢查哪些AJAX請求正在進行。

3）在這里，您可以看到輸入參數，標題和最重要的響應。

4）因此，簡而言之，您可以調試整個AJAX請求生命周期。

Answer 2

您可以從php代碼中回顯json_encode(array('success'=>true))並使用if(data.success){}修改jquery中的if條件if(data.success){}您的修改后的代碼變為

<?php
{
    session_start();
    include_once 'db_connect.php';

    if (isset($_POST))
     {
        $email = filter_input(INPUT_POST, 'email', FILTER_SANITIZE_STRING);
        $logpwd = filter_input(INPUT_POST, 'password', FILTER_SANITIZE_STRING);

        $stmt = $conn->prepare("SELECT password FROM manager WHERE email = ? LIMIT 1");
        $stmt->bind_param("s", $email);


        $stmt->execute();
        $stmt->store_result();

        // get variables from result.
        $stmt->bind_result($password);
        $stmt->fetch();


        // Check if a user has provided the correct password by comparing what they typed with our hash
        if (password_verify($logpwd, $password))
        {
            $sql = "SELECT * from manager WHERE email LIKE '{$email}' LIMIT 1";
            $result = $conn->query($sql);
            $row=mysqli_fetch_array($result);
            $id = $row['id'];
            $conn->query("UPDATE manager SET lastlogin = NOW() WHERE id = $id");

            $_SESSION['manager_check'] = 1;
            $_SESSION['email'] = $row['email'];
            $_SESSION['fullname'] = $row['fullname'];
            $_SESSION['designation'] = $row['designation'];
            $_SESSION['id'] = $row['id'];
            echo json_encode(array('success'=>true));
        }
         else {
            die();
        }
     }
}

和JQuery成為

<script type="text/javascript">
    $(document).ready(function () {
        var form = $('#loginform');
        form.submit(function (ev) {
            ev.preventDefault();
            $.ajax({
                type: form.attr('method'),
                url: form.attr('action'),
                cache: false,
                data: form.serialize(),
                success: function (data) {
                    if (data.success) {
                        $("#user-result").html("<font color ='#006600'> Logged in | Redirecting..</font>").show("fast");
                        setTimeout(
                            function () {
                                window.location.replace("index.php");
                            }, 1500);
                    } else {
                        $("#user-result").html("<font color ='red'> Incorrect Username/Password</font>").show("fast");
                    }

                }

            });
        });
    }); 
</script>