我在查询 2 表专辑和图像中需要一些帮助
我需要从数据库中选择一些字段，它们之间有一个user_id
询问：
$album_query = mysql_query("SELECT `albums`.`album_id`, `albums`.`timestamp`, `albums`.`name`, LEFT(`albums`.`description`, 50) as `description`, COUNT(`images`.`image_id`) as `image_count`
FROM `albums`
LEFT JOIN `images`
ON `albums`.`album_id` = `images`.`album_id`
WHERE `albums`.`user_id` = '**here it must take the session created before**'
GROUP BY `albums`.`album_id`") or die(mysql_error());
这是代码块：
<?php
ob_start();
if(!isset($_SESSION))
{
session_start();
}
require_once('include/connect.php');
if(isset($_GET['user_id']))
{
$id=$_GET['user_id'];
}
elseif(isset($_SESSION['user_id']))
{
$id= $_SESSION['user_id'];
}
else
{
require_once('index.php');
exit();
}
//function to ge the data for the album
function album_data($album_id)
{
}
//check if the album belong to the particular user
function album_check($album_id)
{
}
//get thelist of albums
function get_albums()
{
$albums = array();
$album_query = mysql_query("SELECT `albums`.`album_id`, `albums`.`timestamp`, `albums`.`name`, LEFT(`albums`.`description`, 50) as `description`, COUNT(`images`.`image_id`) as `image_count`
FROM `albums`
LEFT JOIN `images`
ON `albums`.`album_id` = `images`.`album_id`
WHERE `albums`.`user_id` = ''
GROUP BY `albums`.`album_id`") or die(mysql_error());
while($albums_row = mysql_fetch_array($album_query))
{
$albums[] = array(
'id' =>$albums_row['album_id'],
'timestamp'=>$albums_row['timestamp'],
'name'=>$albums_row['name'],
'description'=>$albums_row['description'],
'count'=>$albums_row['image_count']
);
}
return $albums;
}
function create_album($album_name, $album_description)
{
$album_name= mysql_real_escape_string(htmlentities($album_name));
$album_description= mysql_real_escape_string(htmlentities($album_description));
mysql_query("INSERT INTO albums VALUES('', '" .$_SESSION['user_id']. "', UNIX_TIMESTAMP(), '$album_name','$album_description')");
mkdir('uploads/'.mysql_insert_id(), 0744);
mkdir('uploads/thumbs/'.mysql_insert_id(), 0744);
}
?>
*elseif(isset($_SESSION['user_id']))
{
$id= $_SESSION['user_id'];
}*
这是错误，因为系统不允许我分配 album.user_id = $id 并且数据库字段存储 0
我不知道我是否以一种可以理解的方式解释了我的问题
谁能帮我 ？？？