php - 如何使用php在不重新加载/刷新当前页面的情况下显示mysql记录？

Question

场景是这样的:

下面的代码显示所有用户帐户。

<table border="3px" align="center" cellspacing="3px" cellpadding="3px">
<tr>
<td>Username</td>
<td>Password</td>   
<td><a href="" onclick='return popup_link()'>Add</a></td>       
</tr>
<?php
include("connection.php");
$query=mysql_query("select * from tbl_users");
while($row=mysql_fetch_array($query)){
?>
<tr>
<td><?php echo $row['username']; ?></td>
<td><?php echo $row['password']; ?></td>
<td></td>
</tr>
<?php
}
?>
</table>

然后我有一个弹出页面，它可以在不重新加载页面本身的情况下将用户帐户添加到数据库表用户帐户中。

<form id="form">
<h3>Add Record</h3>
<label>Username:</label>
<br/>
<input type="text" id="username" placeholder="Userusername"/>
<br/>
<br/>
<label>Password:</label>
<br/>
<input type="password" id="password" placeholder="Password"/>
<br/>
<br/>
<input type="button" id="submit" value="Submit"/>
</form>


<?php
// Establishing connection with server by passing "server_name", "user_id", "password" 
$connection = mysql_connect("localhost", "root", "");
// Selecting Database by passing "database_name" and above connection variable
$db = mysql_select_db("smsi_inventory", $connection);

//Fetching Values from URL  
$username2=$_POST['username1'];
$password2=$_POST['password1'];

//Insert query 
$query = mysql_query("insert into tbl_users(username, password) values ('$username2','$password2')");
if($query){
echo "Record Saved succesfully";
}

//connection closed
mysql_close($connection);
?>

但是，如果不重新加载/刷新页面，我很难显示新添加的帐户的记录。有没有办法可以使用 php 显示新添加的帐户？

非常感谢您的帮助。谢谢!

Answer 1

尝试执行异步 HTTP (Ajax) 请求。

$.ajax({
    type:'POST',
    url:'YOUR UPDATE DATA URL',
    data:{your:data},
    success:function(data){
      //WHATEVER YOU SHOW ON SUCCESS
    },
    error: function (jqXHR, exception) {
      alert("Parse Errore");
    }
});

更多阅读 http://api.jquery.com/jquery.ajax/