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我想用PHP代碼從我的數據庫中獲取數據。簡言之，當用戶登錄ID爲&密碼我想告訴他唯一的用戶面板數據，但我得到了一些錯誤：這裏是我的用戶管理面板代碼：用戶登錄時從PHP數據庫中獲取具體數據
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/style.css">
<meta charset="utf-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1">
<title>User Panel</title>
<link rel="stylesheet" href="assets/demo.css">
<link rel="stylesheet" href="assets/form-labels-on-top.css">
</head>
<body>
<header>
<h1 align="center"><b>User Information Details</b></h1></br>
</header>
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "xyz";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error)
{
die("Connection failed: " . $conn->connect_error);
}
include('user_login_check.php');
$result= mysql_query("SELECT * FROM `user_information` WHERE `user_id` = '".$_SESSION['id']."' ")or die(mysql_error());
// $result = mysql_query("SELECT * FROM user_information WHERE user_name='" . $_POST["user_name"] . "' and user_password = '". $_POST["user_password"]."'");
?>
<form>
<table border="5" style= "background-color: #333333; color: #FFF; margin: 0 auto; padding:100px" >
<thead>
<tr>
<th>User ID</th>
<th>User Type</th>
<th>User Name</th>
<th>User Country</th>
<th>User Email</th>
<th>User Phone</th>
<td>User Address</td>
<td>User Password</td>
<td>User Status</td>
<td>Edit</td>
</tr>
</thead>
<tbody>
<?php
while($row = mysqli_fetch_assoc($result)){
echo
"<tr>
<td>{$row['user_id']}</td>
<td>{$row['user_type']}</td>
<td>{$row['user_name']}</td>
<td>{$row['user_country']}</td>
<td>{$row['user_email']}</td>
<td>{$row['user_phone']}</td>
<td>{$row['user_address']}</td>
<td>{$row['user_password']}</td>
<td>{$row['user_status']}</td>
<td><a href=\"edit.php?id={$row['user_id']}\"><img src='image/editicon.jpeg'/></a></td>;
</tr>\n";
}
?>
</tbody>
<div align="left" style="background:#333333">
<h1><input type="button" value="Log Out" onClick="window.location.href='logout_user.php'" /> </h1>
</div>
</table>
</form>
</body>
</html>
問題是在專門這一節
$result= mysql_query("SELECT * FROM `user_information` WHERE `user_id` = '".$_SESSION['id']."' ")or die(mysql_error());
這裏是我的登錄校驗碼：
<?php
//connect the database
$hostName="localhost";
$dbUsername="root";
$dbPassword="";
$dbName="xyz";
mysql_connect($hostName,$dbUsername,$dbPassword) or die("Connection failed");
mysql_select_db($dbName) or die("Database name doesn't exist");
//start session
@session_start();
if(isset($_POST["user_name"] , $_POST["user_password"]))
{
$User_name = $_POST["user_name"];
// echo "$User_name";
$User_password = $_POST["user_password"];
// echo "$User_password";
$sql = "SELECT * FROM `user_information` WHERE `user_name`='".$User_name."' AND `user_password`='".$User_password."' ";
$result = mysql_query($sql) or trigger_error(mysql_error().$sql);
// $result = mysql_query("SELECT * FROM `user_info` WHERE `User_name`='".$username."' AND `User_password``='".$password."'");
$my_arary = array();
while($row = mysql_fetch_assoc($result))
{
$_SESSION['id']= $row["Id"];
// echo $_SESSION['login_user_id'];
$_SESSION['user_password']= $row["user_password"];
//echo $_SESSION['login_user_password'];
$_SESSION['user_name']= $row["user_name"];
//echo $_SESSION['login_user_name'];
$my_arary[] = $row;
print_r($row);
if($User_name == $_SESSION['user_name'] && $User_password == $_SESSION['user_password']){
header("Location: userpanel.php");
//echo "successfully logged in";
}
else{
echo "no matches";
header("Location: user_login_basic.php");
}
}
}
?>
顯示你的「user_information」列 – Suhindra
它從數據庫（列）我的用戶信息：\t USER_ID，USER_TYPE，USER_NAME，user_country，USER_EMAIL，USER_PHONE， user_password，user_password – Nayan
在管理面板中添加'session_start（）;' –