1
我在使用此代碼時遇到了一些問題。我正在連接到一個SQL數據庫並且它連接了,但是這個post函數似乎沒有工作。當我檢查元素並查看網絡連接時,它會發佈一個內容,而不是其他兩件事。如果有人能幫助我,那會很棒。Post函數與數據庫無法正常工作
感謝,
本
<?php
$connection = mysql_connect("localhost", "root", "password") or die("There is no connection to the server");
mysql_select_db("tutorial", $connection) or die("Couldn't connect to database");
if ($_POST['login']){
if ($_POST['username'] && $_POST['password']){
$username = mysql_real_escape_string($_POST['username']);
$password = mysql_real_escape_string(hash("sha512", $_POST['password']));
$user = mysql_fetch_array(mysql_query("SELECT * FROM `users` WHERE `Username`='$username'"));
if ($user == '0'){
die("That username does not exist. <a href='index.php'>← Back</a>");
echo "<h1> Test </h1>";
}
if ($user['Password'] != $password) {
die("Incorrect password! <a href='index.php'>← Back</a>");
echo "<h1> Test </h1>";
}
$salt = hash("sha512", rand() . rand() . rand());
setcookie("c_user", hash("sha512", $username), time() + 24 * 60 * 60, "/");
setcookie("c_salt", $salt, time() + 24 * 60 * 60, "/");
$userID = $user['ID'];
mysql_query("UPDATE `users` SET `Salt`='$salt' WHERE `ID`='$userID'");
die("You are now logged in as $username");
}
}
echo "
<body style='font=family: verdana, sans-serif;'>
<div>
<h1>Login to Access Coins</h1>
<br />
<form action='' method='post'>
<table>
<tr>
<td>
<b> Username </b>
</td>
<td>
<input type='text' name='username' style='padding: 6px;' />
</td>
</tr>
<tr>
<td>
<b>Password</b>
</td>
<td>
<input type='password' name='password' style='padding: 6px;' />
</td>
</tr>
<tr>
<td>
<input type='submit' value='Login' />
</td>
</tr>
</table>
</form>
<br />
<h6>
Need an account? <a href='register.php'>Click Here
</h6>
</div>
</body>
";
「郵政功能似乎不工作」是什麼意思? – developerwjk
我爲您介紹一位新朋友,每當您在表單提交時發現錯誤或缺少參數時,它都會對您有所幫助:'print_r($ _ POST);' 您會看到這些值是否正確傳遞到您的頁面。 – technico
@developerwjk:我們必須說「郵政方法似乎沒有工作」,是不是? – technico