Question

我正在开发一个php / MySQL项目，要求我在用户输入用户名时实时检查用户名。

这是我的 username.php ，其中用户实际输入了用户名＆amp;密码以及触发 check.php 的地方......

<html>
<head>
   <script type="text/javascript" src="js/jquery-1.7.1.min.js"></script>

   <script type="text/javascript">

     $(document).ready(function(){
        $("#username").change(function(){
             $("#message").html("<img src='images/loader.gif' /> checking...");


        var username = $("#username").val();

          $.ajax({
                type:"post",
                url:"check.php",
                data:"username =" + username,
                    success:function(data){
                    if(data==0){
                        $("#message").html("<img src='images/tick.png' /><span style='font-size:13px; color: black'> Username available</span>");
                    }
                    else{
                        $("#message").html("<img src='images/err.png' /><span style=font-size:13px; color: red'> Username already taken</span>");
                    }
                }
             });

        });

     });

   </script>
   </head>

   <body>

   <table>
    <tr>
          <td>Username</td>
          <td>:</td>
          <td><input type="text" name="username" id="username"/><td>
            <td id="message"><td>
    </tr>

    <tr>
          <td>Password</td>
          <td>:</td>
          <td><input type="text" name="password" id="password" /><td>
    </tr>
   </table>
   </body>
   </html>

这是 check.php ，其中在数据库中检查用户名。

<?php

$con = mysqli_connect("localhost", "root", "hilbi", "userdb");

if (mysqli_connect_errno())
{

  echo "Failed to connect to MySQL: " . mysqli_connect_error();

}

  // $username = $_POST["username"];

  $username = 'hilbi';

  $query = mysqli_query($con,"SELECT * FROM users WHERE username =   '$username' ");

  $find = mysqli_num_rows($query);

  echo $find;

  mysqli_close($con);

  ?>

现在 check.php 工作正常（我通过单独执行检查，然后返回 1 找到用户名，反之亦然）
但是，当我执行 username.php 时，它始终会返回已使用的用户名 。看起来它实际上并没有在数据库中看到实际上没有这样的用户名。

任何形式的帮助将不胜感激......

Answer 1

使用jQuery在javascript中发送数据：

//some basic validation
if(!isset($_POST['user'] || empty($_POST['user']))
{
   echo '0'; exit();
}

$username = trim($_POST['user']);

在你的check.php中获取这样的用户名

- (void)tableView:(UITableView *)tableView moveRowAtIndexPath:(NSIndexPath *)fromIndexPath toIndexPath:(NSIndexPath *)toIndexPath {
    [queuedToDoArray moveObjectAtIndex:fromIndexPath.row toIndex:toIndexPath.row];

    NSIndexPath *lowerIndexPath = (fromIndexPath.row < toIndexPath.row ? fromIndexPath : toIndexPath);
    NSIndexPath *higherIndexPath = (fromIndexPath.row > toIndexPath.row ? fromIndexPath : toIndexPath);

    //Update all the queue numbers in between moved indexes
    for (int i = lowerIndexPath.row; i <= higherIndexPath.row; i++) {
        NSIndexPath *currentIndexPath = [NSIndexPath indexPathForRow:i inSection:0];
        CustomCell *currentCell = [todoTable cellForRowAtIndexPath:currentIndexPath];
        [currentCell.queueNumber setText:[NSString stringWithFormat:@"%i", currentIndexPath.row + 1]];
    }
}

为了使其正常工作，check.php必须返回1或0，不要在那里或其他任何地回显mysql错误。如果发生任何错误，只需回显0。

Answer 2

您需要正确传递$ username变量。

单引号php变量被视为普通字符串，变量不被评估。

$query = mysqli_query($con,"SELECT * FROM users WHERE username = " .  $username . ";");

OR

$query = mysqli_query($con,"SELECT * FROM users WHERE username = {$username}");

Answer 3

data:"username =" + username

data:"username=" + username

您必须删除用户名和所有工作后的空格

Answer 4

我花了一些时间寻找和测试答案，所以我应该发布我的结果以及对我有用的东西。

html, or php form:

<script src="http://code.jquery.com/jquery-3.3.1.js"></script>
<form class="a16" action="register.php" method="POST" autocomplete="on">
                                    <label for="userName" class=" ">Użytkownik:</label>

<input type="text" name="userName" placeholder="Username" id="userName1" onBlur="checkAvailability()">

      <span id="user-availability-status"></span> 
<p><img src="LoaderIcon.gif" id="loaderIcon" style="display:none" /></p>

 <script type="text/javascript">     

function checkAvailability() {
$("#loaderIcon").show();
jQuery.ajax({
url: "proceser.php",
data:'userName='+$("#userName1").val(),
type: "POST",
success:function(data){
$("#user-availability-status").html(data);
$("#loaderIcon").hide();
},
error:function (){}
});
}
</script>

and proceser.php :

<?php
 include 'dbconnect.php';   //standard datebase local connection..

 if(isset($_POST['userName']) && $_POST['userName']!="") {
     if ($stmt = $con->prepare('SELECT userName FROM users WHERE userName = ?')) {
         $stmt->bind_param('s', $_POST['userName']);
         $stmt->execute();
         $stmt->store_result();
         $numRows = $stmt->num_rows;
         if ($numRows > 0) {
             echo "<span class=''> Username Not Available.</span>";
         } else {
             echo "<span class=''> Username Available.</span>";
         }
     }
 }
$con->close();
ob_end_flush();

?>


Thank you!:)

使用mysql

4 个答案: