我是PHP和Mysql的新手我正在为我的作业编写购物网站,
我的登录,注册和数据库已启动并运行,现在我处于需要执行用户配置文件页面的阶段,该页面需要从数据库中提取数据,
首先我有这个代码,因为它会从我的表中提取所有内容,
<?php
$sql = 'SELECT * from assignment2';
$result = $conn->query($sql);
$rows = $result->fetchAll(PDO::FETCH_ASSOC);
if(count($result)) {
echo '<table><tr>';
//Heading
foreach ($rows[0] as $columnName => $value) {
echo '<th>' . $columnName . '</th>' ;
}
echo '</tr>';
foreach ($rows as $row) {
echo '<tr>';
foreach ($row as $value) {
echo '<td>' . $value . '</td>';
}
echo '<tr>';
}
echo '</table>';
}
?>
但我的问题是如何使它只拉出与登录SESSION匹配的那个
这是我的会话代码
<?php
session_start();
require '../ppuyakul/php/db_conn.php';
if( isset($_SESSION['user_id']) ){
$records = $conn->prepare('SELECT id,username,password FROM assignment2 WHERE id = :id');
$records->bindParam(':id', $_SESSION['user_id']);
$records->execute();
$results = $records->fetch(PDO::FETCH_ASSOC);
$user = NULL;
if( count($results) > 0){
$user = $results;
}
}
?>
提前非常感谢,期待一些不错的答案^^“
答案 0 :(得分:1)
如果id = $ _SESSION [&#39; user_id&#39;]($sql = 'SELECT * from assignment2';
)
$sql = 'SELECT * from assignment2 WHERE id = '.$_SESSION['user_id'];
)
//的foreach
$sql = 'SELECT * from assignment2 WHERE id = '.$_SESSION['user_id'];
$result = $conn->query($sql);
$rows = $result->fetchAll(PDO::FETCH_ASSOC);
if(count($result)) {
echo '<table><tr><th>Id</th><th>Fullname</th><th>Username</th></tr>';
foreach ($rows as $row) {
echo '<tr>';
echo '<td>'.$row["id"].'</td>';
echo '<td>'.$row["fullname"].'</td>';
echo '<td>'.$row["username"].'</td>';
echo '<tr>';
}
echo '</table>';
}