得到數據庫表到其他數據庫表的PHP MySQL的

Question

我想從他的身份證 從數據庫表中獲取用戶名的數據，並把它在其他數據表

function upload_image($image_temp, $image_ext, $album_id, $image_n, $image_description) { 
$album_id = (int)$album_id; 
$image_n = mysql_real_escape_string(htmlentities($image_n)); 
$image_description = mysql_real_escape_string(htmlentities($image_description)); 
//$download_link = 'uploads/'. $album_id. '/'. $image['id']. '.'. $image_ext; 
$mysql_date_now = date("Y-m-d (H:i:s)"); 
$user_name = mysql_query("SELECT `name` FROM `users` WHERE `user_id` = ".$_SESSION['user_id']); 

mysql_query("INSERT INTO `images` VALUES ('', '".$_SESSION['user_id']."','$user_name', '$album_id', UNIX_TIMESTAMP(), '$image_ext', '$image_n', '$image_description', '','$mysql_date_now')"); 


$image_id = mysql_insert_id(); 
$download_link = 'uploads/'. $album_id. '/'. $image_id. '.'. $image_ext; 

mysql_query("UPDATE `images` SET `download_link`='$download_link' WHERE `image_id`=$image_id "); 
$selection = mysql_query("SELECT `user_two` FROM `follow` WHERE `user_one`='".$_SESSION['user_id']."'"); 

while ($row = mysql_fetch_array($selection)) { 
mysql_query("INSERT INTO `notification` VALUES ('', '".$_SESSION['user_id']."', '".$row['user_two']."', '', UNIX_TIMESTAMP(), '$image_n', '$image_description', '$download_link')"); 
} 

$image_file = $image_id.'.'.$image_ext; 
move_uploaded_file($image_temp, 'uploads/'.$album_id.'/'.$image_file); 

Thumbnail('uploads/'.$album_id.'/', $image_file, 'uploads/thumbs/'.$album_id.'/'); 

}

問題這裏

$user_name = mysql_query("SELECT `name` FROM `users` WHERE `user_id` = ".$_SESSION['user_id']); 

mysql_query("INSERT INTO `images` VALUES ('', '".$_SESSION['user_id']."','$user_name', '$album_id', UNIX_TIMESTAMP(), '$image_ext', '$image_n', '$image_description', '','$mysql_date_now')"); 

我得到這個數據庫（資源ID＃14）

Answer 1

您的查詢不返回數據：它返回資源。然後您必須使用資源來檢索您的數據，因此在此行中：

$user_name = mysql_query("SELECT `name` FROM `users` WHERE `user_id` = ".$_SESSION['user_id']); 

$ user_name不包含所需的信息。

嘗試：

$result = mysql_query("SELECT `name` FROM `users` WHERE `user_id` = ".$_SESSION['user_id']); 
list($user_name) = mysql_fetch_array($result); 

注：MySQL是過時 - 使用庫MySQLi或PDO。原理是一樣的。

Answer 2

我想你應該看看這裏： MySQL syntax for Join Update

或者在這裏： MySQL Insert & Joins

您可以用表的一些加盟做到這一點，在短短一個查詢。