AJAX - 检查重复数据并发送通知

Question

我尝试从数据库中检查重复的“用户名”，如果用户名存在，我想使用 AJAX 将消息发送回用户。直到现在，它只插入数据并且验证仍然不起作用。欢迎任何帮助，谢谢！

已编辑：

我正在使用'$.jcrowl' 来获取反馈（参考 add_user.php），当数据插入时它会弹出反馈（例如：'用户成功添加'）。那么如何将这个“在数据库中找到的重复用户名”申请到这个 $.jcrowl 中，我需要如何将验证从“save_user.php”发送回“add_user.php”？

add_user.php

   <div class="row-fluid">
                    <!-- block -->
                    <div class="block">
                        <div class="navbar navbar-inner block-header">
                            <div class="muted pull-left"><i class="icon-plus-sign icon-large"></i> Add Admin User</div>
                        </div>
                        <div class="block-content collapse in">
                            <div class="span12">
                            <form method="post" id="add_user">
                                    <label>First Name :</label>
                                    <input class="input focused" name="firstname" id="focusedInput" type="text" placeholder = "Firstname" required>
                                    <label>Last Name :</label>
                                    <input class="input focused" name="lastname" id="focusedInput" type="text" placeholder = "Lastname" required>
                                    <label>User Type :</label>
                                    <select name="user_type" class="input focused" required/>
                                         <option></option>
                                         <?php $user_level=mysql_query("select * from user_level")or die(mysql_error()); 
                                         while ($row=mysql_fetch_array($user_level)){                                               
                                         ?>
                                         <option value="<?php echo $row['user_type']; ?>"><?php echo $row['type_name']; ?></option>
                                         <?php } ?>
                                    </select>
                                    <label>Username :</label>
                                    <input class="input focused" name="username" id="focusedInput" type="text" placeholder = "Username" required>
                                    <label>Password :</label>
                                    <input class="input focused" name="password" id="focusedInput" type="password" placeholder = "Password" required>

                                        <?php //if admin = 1 and if user = 2
                                        //$session_id=$_SESSION['id'];

                                        $run = $conn->query("select * from users where user_id = '$session_id'")or die(mysql_error());
                                        $user_row = $run->fetch();
                                        $user_type = $user_row['user_type'];


                                        if ($user_type == 1) {
                                        ?>
                                        <div class="control-group">
                                      <div class="controls">


                                        <button  data-placement="right" title="Click to Save" id="save" name="save" class="btn btn-inverse"><i class="icon-save icon-large"></i> Save</button>
                                                <script type="text/javascript">
                                                $(document).ready(function(){
                                                    $('#save').tooltip('show');
                                                    $('#save').tooltip('hide');
                                                });
                                                </script>
                                      </div>
                                    </div>
                                       <?php //not admin

                                            }   
                                        else { ?>
                                            <button  data-placement="right" title="Click to Save" id="save" name="save" class="btn btn-inverse" disabled="disabled"><i class="icon-save icon-large"></i> Save</button> Only admin allowed!
                                                <script type="text/javascript">
                                                $(document).ready(function(){
                                                    $('#save').tooltip('show');
                                                    $('#save').tooltip('hide');
                                                });
                                                </script>





                                        <?php }


                                    ?>
                            </form>
                            </div>
                        </div>
                    </div>
                    <!-- /block -->
                </div>
    <script>
        jQuery(document).ready(function($){
            $("#add_user").submit(function(e){
                e.preventDefault();
                var _this = $(e.target);
                var formData = $(this).serialize();
                $.ajax({
                    type: "POST",
                    url: "save_user.php",
                    data: formData,
                    success: function(html){
                        $.jGrowl("User Successfully  Added", { header: 'User Added' });
                        window.location = 'admin_user.php';  
                    }
                });
            });
        });
        </script>

save_user.php

<?php
include('dbcon.php');
include('session.php');

$firstname = $_POST['firstname'];
$lastname = $_POST['lastname'];
$user_type = $_POST['user_type'];
$username = $_POST['username'];
$password = $_POST['password'];


$query = mysql_query("select * from users where username = '$username' and password = '$password' and firstname = '$firstname' and password = '$password'")or die(mysql_error());
                            $row = mysql_fetch_array($query);
                            $username = $row['username'];

    if ($username == 0) {

    {
        $conn->query("insert into users (username,password,firstname,lastname,user_type) values('$username','$password','$firstname','$lastname','$user_type')")or die(mysql_error());
    }
    else
    {
        echo('USERNAME_EXISTS');
    }
    ?>

Answer 1

首先停止使用 mysql_* 扩展，它在 PHP 7 中已被弃用和关闭。使用 mysqli_* 或 PDO。

解决方案：

您只需在查询中检查用户名，例如：

SELECT * FROM `users` WHERE `username` = '$username'

第二点是，您只需要使用count() 或num rows 函数来检查记录是否存在：

MYSQLi 示例：

$conn = mysqli_connect($dbhost, $dbuser, $dbpass, $db);
$sql = "SELECT * FROM `users` WHERE `username` = '$username'";
$query = mysqli_query($conn,$sql);
$row = mysqli_fetch_array($query);
if(count($row) <= 0){
    //success stuff
}
else{
    // error stuff
}

Answer 2

您只需要检查username 而不是密码。

唯一性条件仅适用于username。

修改查询：

$query = mysql_query("select * from users where username = '$username' and password = '$password' and firstname = '$firstname' and password = '$password'")or die(mysql_error());

收件人：

$query = mysql_query("select * from users where username = '$username' ")or die(mysql_error());

注意：不要使用mysql_* 函数。它们已被弃用，将在未来的 PHP 版本中删除。请改用PDO 或mysqli_*。

Answer 3

我认为您需要在提交后显示错误消息。所以首先您需要在 jquery 中进行一些更改

$.ajax({
data: form_data,
url: "save_user.php",
method: "POST",
dataType: "JSON",
beforeSend: function () {
   // show image if process
}
}).done(function (data) {
if (data.status === "success") {
    $.jGrowl(data.message, { header: 'User Added' });
    window.location = 'admin_user.php';
}
if (data.status === "failure") {
    $.jGrowl(data.message, { header: 'Error Found' });
}
}).error(function () {
$.jGrowl("Some Error Found", { header: 'Error' });
}).complete(function () {

});

在你的 save_user.php 更改这些行

$query = mysql_query("select * from users where username = '$username'");
$number = mysql_num_rows($query);
if ($number == 0) {
    $query2=mysql_query("insert into users (username,password,firstname,lastname,user_type) values
    ('$username','$password','$firstname','$lastname','$user_type')");
    if($query2){
        $data['status'] = "success";
        $data['message'] = "User Created successfully.";
    }else{
        $data['status'] = "failure";
        $data['message'] = "Error: " . mysql_error();
    }

}
else
    {
        $data['status'] = "failure";
        $data['message'] = "USERNAME_EXISTS";
    }
echo json_encode($data);

注意：不要使用 mysql_* 函数。它们已被弃用，将在未来的 PHP 版本中删除。请改用 PDO 或 mysqli_*。