我试图创建一个页面,您可以在其中登录然后更改您的昵称或/和密码。 mySQL 数据库中的所有内容,但是当我尝试将 id 保存到 session 变量时,它不起作用。有什么建议吗?
我使用的是XAMPP,users是我在数据库users中的表,我不会发布登录表单代码,因为它非常简单。
一切都已连接,代码不会给出任何警告或错误。
login.php(片段):
$sql = "SELECT * FROM users WHERE nickname = '$myusername' and pass = '$mypassword' and confirmed = 1";
$result = mysqli_query($conn,$sql);
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
$count = mysqli_num_rows($result);
// If result matched $myusername and $mypassword, table row must be 1 row
if($count == 1) {
$logged = true;
while($row = mysqli_fetch_assoc($result)) {
echo "id: " . $row["id"];
$_SESSION['currentId'] = $row["id"];
echo 'Id: ' . $_SESSION['currentId'];
}
}else {
$error = "Your Login Name or Password is invalid";
}
}
change.php(整个):
<?php
session_start();
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "Users";
$currentId = $_SESSION['currentId'];
if($currentId<1){echo 'No Id.';}
else {echo 'CurrentId: ';
echo $currentId;}
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
echo "Connected successfully <br>";
if($_SERVER["REQUEST_METHOD"] == "POST") {
// username and password sent from form
$aCUname = mysqli_real_escape_string($conn,$_POST['CUname']);
$aCUpass = mysqli_real_escape_string($conn,$_POST['CUpass']);
$sql = "UPDATE users SET nickname = '$aCUname', pass = '$aCUpass' WHERE id = '$currentId';";
$result = mysqli_query($conn,$sql);
echo 'Updated successfully.';
}
?>
感谢您的帮助。
最佳答案
我找到了解决方案。我只得删除
$row = mysqli_fetch_array($result,MYSQLI_ASSOC);
来自login.php。感谢@Shashikumar Misal!
关于Php 获取当前用户的 id 不起作用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/47580937/