尝试从数据库中的列中减去金额，显示所有帐户的金额

Question

目标：

我试图通过按下 html 按钮从我的数据库中的列中减去 0.05。

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出了什么问题：

一切正常，它进行了正确的计算并从我的帐户中删除了正确的金额，但它在数据库中的每个帐户上显示了新的金额。

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示例：

我的帐户中有 10 个，一旦我按下按钮，我的帐户中现在就有 9.95 个。不幸的是，现在每个人的账户中也都有 9.95！

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代码：

Html 按钮和 onclick 脚本：

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

    <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>

减0.05的PHP代码（文件名：subtract5.php）：

<?php
session_start();

$servername = "localhost";
$username = "my username"; <not actually this, just confidential
$password = "my password"; <not actually this, just confidential
$dbname = "accounts";
$cash_amount = $_SESSION['cash_amount'];

// Create connection

$userid = $_SESSION['id'];

// 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 = $cash_amount - 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();
?>

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预测：

我认为这与用户 ID 字段有关，我不知道该放什么。

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数据库布局： 数据库：帐户， 表：用户， 列：id |名字 |姓氏 |电子邮件 |密码 |现金金额 |哈希 |活跃

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额外问题：

在我网站的主页上，它显示了现金金额。一旦它被更改，它不会更新，直到我注销并重新登录。有没有什么我可以在顶部的 php 代码中放入诸如“会话重启”之类的东西，以便在每次打开页面时检查新的金额？

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谢谢：

非常感谢你的帮助，你可以看到我在这方面有点菜鸟:)

Answer 1

在查询的where子句中WHERE id = $userid:

<?php
session_start();

$servername = "localhost";
$username = "my username"; <not actually this, just confidential
$password = "my password"; <not actually this, just confidential
$dbname = "accounts";
$cash_amount = $_SESSION['cash_amount'];

// Create connection

$userid = $_SESSION['id'];

// 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 = $cash_amount - 0.05;

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

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

$conn->close();
?>

Answer 2

您需要使用 WHERE 子句将更新限制为仅限一个用户

$sql = "UPDATE users SET cash_amount = $newAmount WHERE user_id = $userid LIMIT 1";

添加 LIMIT 是为了安全。