PHP代码运行时不单击按钮_Php_Database

PHP代码运行时不单击按钮

php database

PHP代码运行时不单击按钮,php,database,Php,Database,我在一个使用PHP的网站上工作，我想在点击按钮时更新数据库。但出于某种原因，无论何时加载页面，代码都会运行，我不希望这样，因为它可能会真正打乱整个代码。如何停止脚本自动运行 <?php ob_start(); session_start(); include_once 'dbconnect.php'; // if session is not set this will redirect to login page if( !isset($_SE

我在一个使用PHP的网站上工作，我想在点击按钮时更新数据库。但出于某种原因，无论何时加载页面，代码都会运行，我不希望这样，因为它可能会真正打乱整个代码。如何停止脚本自动运行

<?php
    ob_start();
    session_start();
    include_once 'dbconnect.php';


    // if session is not set this will redirect to login page
    if( !isset($_SESSION['user']) ) {
        header("Location: index.php");
        exit;
    }

    $res=mysql_query("SELECT * FROM users WHERE userId=".$_SESSION['user']);
    $userRow=mysql_fetch_array($res);
    //Here is where the script is
    if ( isset($_POST['send']) ) {
        if ( ! empty($_POST['sender'])){
            $name = $_POST['sender'];
        }
        if ( ! empty($_POST['reciever'])){
            $name = $_POST['reciever'];
        }

        $query = "UPDATE users SET userCoins = userCoins + 1  WHERE userName='Morgan'";
        $res = mysql_query($query);
        if ($res) {
            $error = "Success!";
        } else {
            $error = "Something Went Wrong!";
        }
    }
?>
<!DOCTYPE html>
<html>
    <?php header("Access-Control-Allow-Origin: http://www.py69.esy.es"); ?>
    <head>
        <title>ServiceCoin</title>
        <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
        <link rel="stylesheet" href="assets/css/bootstrap.min.css" type="text/css"  />
        <link rel="stylesheet" href="scripts/home/index.css" />
    </head>
    <body>
        <ul>
            <li><a href="#" class="a">ServiceCoin.com(image)</a></li>
            <li><a href="logout.php?logout" class="a">Sign Out</a></li>
            <li><a href="#" class="a">Contact</a></li>
            <li><a href="#" class="a">Get Service Coins</a></li>
            <li><a href="#" class="a">News</a></li>
            <li><a href="settings.php" class="a">Settings</a></li>
            <li><a href="#" class="a">Referrals</a></li>
            <li><a href="service.php" class="a">Services</a></li>
            <li><a href="home.php" class="a">Home</a></li>
        </ul>
        <br /><br />
        <center>
        <h3>Welcome, <?php echo $userRow['userName']; ?>. You Currently Have <span id="services"><?php echo $userRow['userCoins']; ?></span> Service Coins</h3>
        <form method="post" action="<?php echo htmlspecialchars($_SERVER['PHP_SELF']); ?>" autocomplete="off">
            <div class="form-group">
                <div class="input-group">
                    <span class="input-group-addon"><span class="glyphicons glyphicons-lock"></span></span>
                    <input type="text" name="sender" class="form-control" placeholder="Enter Your Wallet Key" value="<?php echo $row['userCoins']; ?>" maxlength="15" />
                    <span class="text-danger"><?php echo $error; ?></span>
                </div>
                <div class="input-group">
                    <span class="input-group-addon"><span class="glyphicons glyphicons-lock"></span></span>
                    <input type="text" name="reciever" class="form-control" placeholder="Enter The Recievers Wallet Key" value="<?php echo $row['userCoins']; ?>" maxlength="15" />
                    <span class="text-danger"><?php echo $error; ?></span>
                </div>

            </div>
            <div class="form-group">
                <button type="submit" class="btn btn-block btn-primary" name="send">Sign Up</button>
            </div>
        </form>
        </center>
    </body>
</html>
<?php ob_end_flush(); ?>

。您目前有服务币
. 您目前有服务币
您应该使用MySQLi
。关于为什么不mysql.*
您可以阅读
解决方案
您的代码可能如下所示：
$mysqli = new mysqli("localhost", "my_user", "my_password", "table_name"); // here you will need your connection data, you can store it in dbconnect.php.

// if session is not set this will redirect to login page    
if(!isset($_SESSION['user'])) {
    header("Location: index.php");
    exit;
}

$condition = empty($_POST['sender']) || empty($_POST['reciever']);
if (!$condition) {
    $res= "SELECT * FROM users WHERE userId=".$_SESSION['user'];
    $mysqli->query($res); // you are doing nothing with it in your code, why?
    $name = $_POST['sender'];       
    $reciever = $_POST['reciever'];

    $query = "UPDATE users SET userCoins = userCoins + 1  WHERE userName='Morgan'";
    $res = $mysqli->query($query);
    if ($res) {
        $error = "Success!";
    } else {
        $error = "Something Went Wrong!";
        echo "Error: ".$mysqli->error; // here you can check your errors
    }
}

手册
有关MySQLi
的更多信息，请参见die（）
将停止脚本。另一方面，我认为mysql.*
可能会真正打乱您的整个代码…我将如何在这方面使用die方法？@Rasclatt您对此有什么改进吗，任何东西都很受欢迎，因为我是PHPI新手，无法发现任何明显的原因，但解决它的方法是基本的调试-设置一些断点并逐步完成代码。如果当前没有调试器，请安装Xdebug。不完全是问题，但$error=“Success！”让我发笑…我在$condition行上得到一个错误，它说有一个意外的“；”我仍然在同一行上得到错误，我不知道为什么它看起来讨厌它。现在：D我已经做了一些语法，对不起；）我添加了一些echo当查询失败时，您可以使用它进行调试。。。我的页面现在完全是白色的。我将更新问题中的代码
$mysqli = new mysqli("localhost", "my_user", "my_password", "table_name"); // here you will need your connection data, you can store it in dbconnect.php.

// if session is not set this will redirect to login page    
if(!isset($_SESSION['user'])) {
    header("Location: index.php");
    exit;
}

$condition = empty($_POST['sender']) || empty($_POST['reciever']);
if (!$condition) {
    $res= "SELECT * FROM users WHERE userId=".$_SESSION['user'];
    $mysqli->query($res); // you are doing nothing with it in your code, why?
    $name = $_POST['sender'];       
    $reciever = $_POST['reciever'];

    $query = "UPDATE users SET userCoins = userCoins + 1  WHERE userName='Morgan'";
    $res = $mysqli->query($query);
    if ($res) {
        $error = "Success!";
    } else {
        $error = "Something Went Wrong!";
        echo "Error: ".$mysqli->error; // here you can check your errors
    }
}