php - MYSQL实时显示

Question

我希望我的网站显示每个用户的硬币，当我向用户添加一些硬币时，我希望他不需要重新加载页面即可获得正确的金额。

这么短，我只想实时显示 mysql 值。

我的代码

                <?php
                if($admin == 1)
				{
				echo'
				<li class="text-muted menu-title">Admin Panel</li>
				<li class="has_sub admin-content">
                <a href="#" class="waves-effect '; if($page=="ad"){echo'subdrop';} echo'"><i class="ti-user"></i><span>Admin</span> </a>
                <ul class="list-unstyled '; if($page=="ad"){echo'"style="display: block;"';} echo'" style="">
				<li><a href="apu.php">Premium users</a></li>
				<li><a href="abu.php">Banned users</a></li>
				<li><a href="au.php">Users</a></li>
                </ul>
				</li>
				';
                         
    }
	if(isset($_GET["action"]))
{
	if($_GET["action"] == "view")
	{
		$sid = $_GET["id"];
	}
	if($_GET["action"] != "view")
	{
		echo '<script>location.href="index.php" </script>';
	}

}
if(!isset($_GET["action"]))
{
	if(isset($_SESSION["steamid"]))
	{
		$sid= $_SESSION['steamid'];
	}
	else
	{
		echo '<script>location.href="index.php" </script>';
	}
}

$sid=mysql_real_escape_string($sid);

$exists=fetchinfo("steamid","users","steamid",$sid);

if(!$exists)
{
	echo '<script>location.href="index.php" </script>';
}

	$crdts=fetchinfo("credits","users","steamid",$sid);	

	
	if($reg)
	{
		$reg2date=date('Y-m-d', $reg);
	}
	else
	{
		$reg2date='Unknown';
	}
	if($gp==0  || $gw==0)
	{
		$wr=0;
	}
	else
	{
		$wr=($gw/$gp)*100;
	}
	
	if($premium==1)
	{
	$id=$_SESSION['steamid'];
	$time=time();
	$puntil = fetchinfo("puntil","users","steamid","$id");
	if($puntil<=$time)
	{
		
		mysql_query("UPDATE users SET `premium`='0' WHERE `steamid`='$id'");
		mysql_query("UPDATE users SET `profile`='1' WHERE `steamid`='$id'");
		
	}
}

?>
		
		
									    <div class="credits1">
                                        <li class="text-muted menu-title">Credits</li>
                                        <br>
										</div>
										<div id="credits"><?php echo $crdts; ?></div>
                       

感谢您的关注，希望您能帮助我。

Answer 1

您需要使用ajax请求从源获取异步数据。最常见的方式是javascript ajax请求，但我假设你已经了解jquery，如果不知道你可以快速了解它，请参阅:http://jquery.com

首先您需要设置一个 php 文件，该文件从您需要的数据库返回数据

假设ajax.php

// make db connection
// get related data from db
// return the data (coins)

// set result array
$result = [
   'coins' => $dbResult['coins']
];

// you can use json, so that you can easily manipulate it
// encode to json the array
$json = json_encode($result);

echo $json;

// or you could jus echo the coins as well, like:
echo $dbResult['coins'];

ajax.js

$(function(){
    $.ajax({
      url: "/api/getWeather",
      dataType: json, // if you get a json, using this better than parsing result to json
      success: function( result ) {
       // updating coins in the DOM for json return
       $( "selector to show coins" ).text(result.coins);

       // or updating coins for plain text return
       $( "selector to show coins" ).text(result);
      }
    });
});

您也可以更新代码或向代码添加新功能。