Question

我试图在按下按钮后从数据库中的玩家现金金额减去0.05。这是我到目前为止所得到的。

我的数据库：

数据库名称：帐户

表：用户

我想影响的专栏： cash_amount

HTML：

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.3/jquery.min.js"></script>
<script type = "text/javascript">
function myAjax () {
$.ajax( { type : 'POST',
            data : { },
            url  : 'subtract5.php',              // <=== CALL THE PHP FUNCTION HERE.
            success: function ( data ) {
                alert( data );               // <=== VALUE RETURNED FROM FUNCTION.
            },
            error: function ( xhr ) {
                alert( "error" );
            }
        });
}
</script>


<button id="playbutton" onclick="myAjax();location.href='5game.html'">Play (-5&#162;)</button>

Php档案:( subtract5.php）

<?php

 UPDATE `accounts`.`users` SET `cash_amount` = '`cash_amount` - 0.05'

感谢您的帮助，我有点像菜鸟。）

Answer 1

以下代码，但是不要完全复制粘贴它，读取注释有一些变量，你可能需要更改并从数据库中获取。

<?php
$servername = "servarname";
$username = "username";
$password = "password";
$dbname = "dbName";

// Create connection

$userid = // You must enter the user's id here.

$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection

if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

// Fetch the existing value of the cash_amount against that particular user here. You can use the SELECT cash_amount from users where userid = $userid

$newAmount = $previousAmount - 0.05;

$sql = "UPDATE users SET cash_amount = '$newAmount'";
$result = $conn->query($sql);

if($result)
{
   echo "Query Executed Successfully!";
}
else
{
   echo mysqli_error($conn);
}

$conn->close();
?>

Answer 2

尝试执行该请求。这只不过是一个没有引号的字符串。我确信PHP认为它是一个错误。

在尝试编写项目代码之前，您应该考虑学习PHP和MySQL。

按html按钮后如何从mysql数据库中减去？

2 个答案: