PHP文件没有在AJAX中运行？

我的AJAX函数没有运行我的PHP脚本。我有一个类似的功能，我的注册，它是没有任何问题的工作。AJAX函数运行到done调用的末尾，但是PHP文件中创建cookie的第一行没有运行

我已经将数据（结果）输出到控制台，看起来一切正常。没有出现错误，我已经在我完全控制的服务器上启用了PHP错误。服务器很好，因为网站的注册部分使用了这种精确的方法，并且它工作正常

JQuery

$('#login-form').on("submit", function(e){
    var dataString = $(this).serialize();
    console.log(dataString);
    $.ajax({
        type: "POST",
        url: "bin/login.php",
        data: dataString
    }).done(function (result, status, xhr) {
            // Display message back to the user here.
            window.location.replace('./app/');
            console.log("Login Completed!");
        }).fail(function (result) {
            // TASK: Add a visual element on the page for errors.
            if (result.status == 522){
                alert("Email not verified!");
            } else if (result.status == 523){
                alert("Password was incorrect!");
            } else if (result.status == 524) {
                alert("User account not found!");
            }
        });
    return false;
});

PHP

<?php
    setcookie("TestCookie", "login.php", none, "/");
    if (array_key_exists('emailAddress', $_POST) && array_key_exists('password', $_POST)){
        $emailAddress = $_POST["emailAddress"];
        $password = $_POST["password"];
        // Connect to the database with this standard connect PHP script.
        include('connectDB.php');
        // Check the link was established.
        if (!mysqli_connect_error()) {
            $sqlget = "SELECT id, emailAddress, userPassword, userRole, emailVerified FROM users WHERE emailAddress=\"$emailAddress\"";
            $sqldata = mysqli_query($dbconn, $sqlget);
            
            if (mysqli_num_rows($sqldata) == 1){                
                // One user was found.
                $userInfo = mysqli_fetch_array($sqldata);                
                setcookie("Account", "found", none, "/");
                if ((password_verify($password, $userInfo["userPassword"])) {   
                    setcookie("Password", "OK", none, "/");
                    if ($userInfo["emailVerified"] == true) {
                        setcookie("loginId", $userInfo["id"], none, "/");
                    } else {
                        // Email was not verified.
                        header('HTTP/1.1 522 Email Not Verified');
                        header('Conent-Type: application/json; charset=UTF-8');
                        die(json_encode($result));                                    
                    }                    
                } else {
                    // Password verification failed.
                    header('HTTP/1.1 523 Password verification Failed');
                    header('Conent-Type: application/json; charset=UTF-8');
                    die(json_encode($result));   
                }
            } else {
                // No user found in the system
                header('HTTP/1.1 524 User not Found');
                header('Conent-Type: application/json; charset=UTF-8');
                die(json_encode($result));    
            }
        } else {
            // Place code to tell the user there was an internal DB error.
            // Possibly a standard error message.  Lets not scare the user.        
            header('HTTP/1.1 500 Internal Server Error');
            header('Conent-Type: application/json; charset=UTF-8');
            die(json_encode($result));
        }
    }    
?>

---

我发现了问题。我在第18行的括号太多了。我将行更改为以下内容，并运行PHP脚本

if (password_verify($password, $userInfo["userPassword"])) {
    if ($userInfo["emailVerified"] == true) {
        setcookie("loginId", $userInfo["id"], 0, "/");
    } else {
        // Email was not verified.
        header('HTTP/1.1 522 Email Not Verified');
        header('Conent-Type: application/json; charset=UTF-8');
        die(json_encode($result));                                    
    }                    
} else {
    // Password verification failed.
    header('HTTP/1.1 523 Password verification Failed');
    header('Conent-Type: application/json; charset=UTF-8');
    die(json_encode($result));   
}

您的
setcookie（“TestCookie”、“login.php”、none“/”）中的
none
是什么
不依赖javascript来读取cookie，我只需要依靠PHP来验证他们是否登录了etc，并将错误消息返回给javascript。如果不希望cookie过期，请将
none
更改为零（0）。@imvain2是一个更好的方法。谢谢你引起我的注意。尽管如此，仍然无法解释为什么PHP文件没有首先运行。@Kisaragi，谢谢。我确实看到了一条消息，没有人贬值。一旦我解决了这个问题，我就有一张便条要查。谢谢你为我节省时间。我更改了，但代码仍然无法运行。