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在我的網站我有我想要的是能夠在數據庫中更新信息的管理頁面,使用表格。使用PHP更新一個庫MySQLi數據庫和HTML5形成
這是一個利用輸入信息和更新什麼是我的數據庫中的代碼IM:
adminform.php
<html>
<head>
<link rel="stylesheet" href="assets/css/main.css" />
</head>
<body>
<header id="header">
<h1><a href="home.php">SafeTNet</a></h1>
<nav id="nav">
<ul>
<li>Admin Page Only</li>
<li></li>
<li><a href="logout.php" class="button">Logout</a> </li>
</ul>
</nav>
</header>
<h1> Select a member </h1>
<br />
<select name="members" onchange="showUser(this.value)">
<option value="">Select a member email</option>
<?php
$query = "SELECT * FROM members";
$mysqli = new mysqli('localhost','root','root','SafeTNetD');
$result = $mysqli->query($query);
while($row = $result->fetch_assoc())
echo '<option value="'.$row["email"].'">'.$row["email"].'</option>';
?>
</select>
<div id="signup">
<h2>Update Your Member Information</h2>
<form method="post" action="admin1.php">
<table>
<tr>
<td>Email</td>
<td><input type="text" name="email" required="required"></td>
</tr>
<tr>
</tr>
<tr>
<td>City </td>
<td><input type="text" name="city"></td>
</tr>
<tr>
</tr>
<tr>
</table>
<br><br>
<div id="buttons">
<input type="submit">
</div>
</body>
</html>
admin1.php
<html>
<head>
<title>Admin</title>
<link rel="stylesheet" href="assets/css/main.css" />
</head>
<body>
<header id="header">
<h1><a href="home.php">SafeTNet</a></h1>
<nav id="nav">
<ul>
<li>Admin Page Only</li>
<li></li>
<li><a href="logout.php" class="button">Logout</a></li>
</ul>
</nav>
</header>
<br />
<?php
$query = "SELECT * FROM members";
$mysqli = new mysqli('localhost','root','root','SafeTNetD');
$result = $mysqli->query($query);
while($row = $result->fetch_assoc())
echo '<option value="'.$row["email"].'">'.$row["email"].'</option>';
?>
</select>
<br />
<?php
$q=$row["email"];
$mysqli = new mysqli('localhost','root','root','members');
$sql = "SELECT * FROM members WHERE email='".$q."'";
if(array_key_exists('_submit_check', $_POST))
{
$email = $_POST['email'];
$city = $_POST['city'];
$sql = "UPDATE members SET city = '$city' WHERE email = '$q'";
if($mysqli->query($sql) === TRUE)
{
echo 'Record updated successfully<br />';
}
else
{
echo $sql.'<br />' . $mysqli->error;
}
$mysqli->close();
}
?>
<br><br><br>
<footer id="footer">
<img src="logo.jpg" height="50px">
<ul class="copyright">
<li>© SafeTNet. All rights reserved.</li><li> 2016</li>
</ul>
</footer>
</body>
</html>
我CA n得到表單運行,但不能獲取信息在數據庫中更改或回顯到屏幕。
預先感謝您。