Question

我的网站有一个奇怪的问题。经过整整2天的努力解决后，我仍然没有设法找到错误。

我最近注意到（我认为它是在更改服务器之后开始的）我的ajax代码不再正常工作了。在它像魅力一样工作之前。以下是我在线社区的登录脚本示例：

遵循表格：

<form class="registration-form" name="signupform" onsubmit="return false;"> 
     <input type="text" onfocus="emptyElement('status')" placeholder="Email"   id="email" autofocus>
     <input type="password" onfocus="emptyElement('status')" placeholder="Password" id="password">
     <button type="submit" id="loginbtn" onclick="login()" >Log in!</button>
     <div id="status"></div>
</form>

关注JS：

function emptyElement(x){
    _(x).innerHTML = "";
}
function login(){
    var e = _("email").value;
    var p = _("password").value;
    if(e == "" || p == ""){
        _("status").innerHTML = 'Please fill out all the form data.';
    } else {
        _("loginbtn").style.display = "none";
        _("status").innerHTML = 'Loading...';
        var ajax = ajaxObj("POST", "login.php");
        ajax.onreadystatechange = function() {
            if(ajaxReturn(ajax) == true) {
                if(ajax.responseText == "login_failed"){
                    _("status").innerHTML = 'Login unscuccessful';
                    _("loginbtn").style.display = "block";
                } else {
                    window.location = "xxxx";
                }
            }
        }
        ajax.send("e="+e+"&p="+p);
    }
}

关注PHP：

if(isset($_POST["e"])){
    include_once("php_includes/db_conx.php");
    $e = mysqli_real_escape_string($db_conx, $_POST['e']);
    $e = strtolower($e);
    $p = md5($_POST['p']);
    $ip = preg_replace('#[^0-9.]#', '', getenv('REMOTE_ADDR'));
    if($e == "" || $p == ""){
        echo "login_failed";
        exit();
    } else {
        $query = mysqli_query($db_conx,"SELECT id, username, password FROM users WHERE email='$e' AND activated='1' LIMIT 1");
        $row = mysqli_fetch_row($query);
        $db_id = $row[0];
        $db_username = $row[1];
        $db_pass_str = $row[2];
        if($p != $db_pass_str){
            echo "login_failed";
            exit();
        } else {
            //set cookies and log user in
        }
    }
    exit();
}

虽然表单的php正在运行，但当用户登录时所有值都正确时，ajax没有显示任何状态或加载消息，就像上面的代码一样。< / p>

脚本没有其他任何改变。我仍然像以前一样在页面顶部添加ajax.js和jquery。

这只是一个例子。我用来让人们注册的表单和这个表单一样，并且有同样的问题，就像我网站上的任何其他ajax表单一样。

问题是什么？

Answer 1

在从服务器获取数据到javascript（AJAX）时更好地使用json

if(isset($_POST["e"])){
    include_once("php_includes/db_conx.php");
    $e = mysqli_real_escape_string($db_conx, $_POST['e']);
    $e = strtolower($e);
    $p = md5($_POST['p']);
    $ip = preg_replace('#[^0-9.]#', '', getenv('REMOTE_ADDR'));

    $json['success'] = true;

    if($e == "" || $p == ""){
        $json['success'] = false;
    } else {
        $query = mysqli_query($db_conx,"SELECT id, username, password FROM users WHERE email='$e' AND activated='1' LIMIT 1");
        $row = mysqli_fetch_row($query);
        $db_id = $row[0];
        $db_username = $row[1];
        $db_pass_str = $row[2];
        if($p != $db_pass_str){
            $json['success'] = false;
        } else {
            //set cookies and log user in
        }
    }
    // display your return data as json
    echo json_encode($json);
}

将onreadystatechange函数的内部包装到

if (this.readyState == 4 && this.status == 200) {

}

删除您的条件

if(ajaxReturn(ajax) == true) {
}

您的js文件现在就像这样

function emptyElement(x){
    _(x).innerHTML = "";
}
function login(){
    var e = _("email").value;
    var p = _("password").value;
    if(e == "" || p == ""){
        _("status").innerHTML = 'Please fill out all the form data.';
    } else {
        _("loginbtn").style.display = "none";
        _("status").innerHTML = 'Loading...';
        var ajax = ajaxObj("POST", "login.php");
        ajax.onreadystatechange = function() {
            // when ajax is now okay and request finished and response is ready
            if (this.readyState == 4 && this.status == 200) {
                var result = JSON.parse(this.responseText);

                // debugging purpose display response
                console.log(result);

                if(result.success == true){
                    _("status").innerHTML = 'Login unscuccessful';
                    _("loginbtn").style.display = "block";
                } else {
                    window.location = "xxxx";
                }
            }
        }
        ajax.send("e="+e+"&p="+p);
    }
}

Ajax神秘地停止在我的整个网站上工作

1 个答案: