我正在创建一个注册页面,用户只能在其中选择可用的用户名,即数据库中不存在的用户名.所以我为此使用ajax ..但是当我一次输入太多请求时,它失败了,我的网站停止工作..一段时间后,一切都会好起来的,但5-10分钟后它将停止工作.

网站提供错误：-

此页面无法正常运行.collegespike.com没有发送任何数据.
ERR_EMPTY_RESPONSE

从数据库获取数据的代码：

<?php

if (isset($_POST['user_name']) && $_POST['user_name'] != '') {
    $response = array();
    $username = mysqli_real_escape_string($db, $_POST['user_name']);
    $sql = "select username from login_master where username='" . $username . "'";
    $res = mysqli_query($db, $sql);
    $count = mysqli_num_rows($res);
    if ($count > 0) {
        $response['status'] = false;
        $response['msg'] = 'Username already exists.';
    } else if (strlen($username) < 6 || strlen($username) > 15) {
        $response['status'] = false;
        $response['msg'] = 'Username must be 6 to 15 characters';
    } else if (!preg_match("/^[a-zA-Z1-9]+$/", $username)) {
        $response['status'] = false;
        $response['msg'] = 'Use alphabet and numeric only.';
    } else {
        $response['status'] = true;
        $response['msg'] = 'Username is available.';
    }

    echo json_encode($response);
}
?>

Ajax代码：-

//including jquery 2.0.3
<script type="text/javascript" charset="utf8" src="https.://ajax.aspnetcdn.com/ajax/jQuery/jquery-2.0.3.js"></script>

//jquery code to fetch data
<script type="text/javascript">
$(document).ready(function() {
    $('#username').keyup(function() {
        var usercheck = $(this).val();
        $('#usercheck').html('<img src="images/loading.gif" width="150" />');
        $.post("signup/check1.php", {
            user_name: usercheck
        }, function(data) {
            if (data.status == true) {
                $('#usercheck')
                   .parent('div')
                   .removeClass('has-error')
                   .addClass('has-success');

            } else {
                $('#usercheck')
                   .parent('div')
                   .removeClass('has-success')
                   .addClass('has-error');
            }
            $('#usercheck').html(data.msg);
        }, 'json');
    });
});
</script>

HTML代码：
//代码以接受输入并显示输出

//taking input
<input class="check-form"  name="user_name" id="username" type="text"  placeholder="User Name (please enter unique username)">

//showing output
<span id="usercheck" class="help-block"></span>
<span><i class="ti-check"></i></span>

解决方法:

我认为问题是由于托管服务器设置而来的.如果您触发过多的发布请求,则服务器会因设置而阻止您60到120秒.

尝试使用GET请求而不是POST请求,我认为它将解决问题.

$(document).ready(function() {
    $('#username').keyup(function() {
        var usercheck = $(this).val();
        $('#usercheck').html('<img src="images/loading.gif" width="150" />');
        $.get("signup/check1.php", {
            user_name: usercheck
        }, function(data) {
            if (data.status == true) {
                $('#usercheck')
                   .parent('div')
                   .removeClass('has-error')
                   .addClass('has-success');

            } else {
                $('#usercheck')
                   .parent('div')
                   .removeClass('has-success')
                   .addClass('has-error');
            }
            $('#usercheck').html(data.msg);
        }, 'json');
    });
});