我的代码出现问题，用户名和密码不匹配。
这是我的输出。当用户名和密码不匹配时，系统会提示用户名和密码不匹配不匹配，但是，当用户名和密码匹配时，就会出现一条消息，表明它是match下面是我的代码：
html代码
<body>
<div class="container box">
<div class="form-group">
<h3 align="center">Live Username Available or not By using PHP Ajax Jquery</h3><br />
<label>Enter Username</label>
<input type="text" name="username" id="username" class="form-control" />
<input type="password" name="password" id="password" class="form-control" />
<span id="availability"></span>
<br /><br />
<button type="button" name="register" class="btn btn-info" id="register" disabled>Register</button>
<br />
</div>
<br />
<br />
</div>
</body>
</html>
脚本代码
<script>
$(document).ready(function(){
$('#username','#password').blur(function(){
var username = $(this).val();
var password = $(this).val();
$.ajax({
url:'check.php',
method:"POST",
data:{user_name:username, password:password},
success:function(data)
{
if(data != '1')
{
$('#availability').html('<span class="text-danger">Username and Password not Match</span>');
$('#register').attr("disabled", true);
}
else
{
$('#availability').html('<span class="text-success">Username and Password Available</span>');
$('#register').attr("disabled", false);
}
}
})
});
});
</script>
检查.php- 我的数据库连接和从数据库获取它的查询
<?php
//check.php
$connect = mysqli_connect("localhost", "username", "", "dbname");
if(isset($_POST["user_name"] && $_POST["password"]))
{
$username = mysqli_real_escape_string($connect, $_POST["user_name"]);
$password = mysqli_real_escape_string($connect, $_POST["password"]);
$query = "SELECT * FROM admin WHERE username = '".$username."' AND password = '".$password."' ";
$result = mysqli_query($connect, $query);
echo mysqli_num_rows($result);
}
?>
