I am building a commenting system where people can comment on uploaded files, messages and to-do items. What is the best way to connect the comment table table to the other various tables?
Possible Solutions
Solution one - use a two field foreign key.

CREATE TABLE `comments`(
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
foreign_key INT NOT NULL,
table_name enum('files','messages','to-do'),
user_id INT NOT NULL,
comment TEXT NOT NULL);

Solution two - Her tablo veritabanına benzersiz bir birincil anahtar olurdu. Yani her tablo için birincil anahtar olarak PHP'nin uniqid ($ prefix) kullanmak istiyorsunuz.

CREATE TABLE `comments`(
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
foreign_key char(23) NOT NULL,
table_name enum('files','messages','to-do'),
user_id INT NOT NULL,
comment TEXT NOT NULL);

Solution Three - Yorum tablodaki birden fazla yabancı anahtarları var

CREATE TABLE `comments`(
id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
files_id INT NOT NULL,
messages_id INT NOT NULL,
to_do_id INT NOT NULL,
user_id INT NOT NULL,
comment TEXT NOT NULL);

En iyi çözüm nedir? Ben giriş takdir ve ben bir şey açıklamak eğer lütfen bana bildirin

EDIT removed table_name from solution three as it was a copy_paste error As to Joe's Response

Assume: 1) all data is already escaped. Do we really need to see that?
2) $fileId = "146".
3) $userId = "432".
4) $comment = "Stackoverflow is so awesome!"

INSERT

$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
 mysql_select_db('mydb');
 mysql_query("INSERT INTO `comments` (user_id,comment) VALUES($userId,$comment)");
 $commentId = mysql_insert_id();
 mysql_query("INSERT INTO `comments_files_xref` (file_id,comment_id)         VALUES($fileId,$commentId)");