Diğer yanıtlar belirtildiği gibi, bir kendi tablo olarak temsil kullanıcıları ve etiketleri arasında olan bir çok-çok ilişkisi, büyük olasılıkla var. Ben basitleştirilmiş bir durumda bir SQL demo yaptı. InterestingTags tablo ne etiketleri ilgi ne kullanıcı bağlayan tablodur.
/* Create tables */
CREATE TABLE User (id INT NOT NULL AUTO_INCREMENT, name varchar(50), PRIMARY KEY(id));
CREATE TABLE Tag (id INT NOT NULL AUTO_INCREMENT, name varchar(50), PRIMARY KEY(id));
CREATE TABLE InterestingTags (user_id INT NOT NULL REFERENCES User(id), tag_id INT NOT NULL REFERENCES Tag(id), PRIMARY KEY(user_id,tag_id));
/* Insert some data */
/* 3 users, 5 tags and some connections between users and tags */
INSERT INTO User (name) VALUES ('jQueryFreak'), ('noFavoriteMan'), ('generalist');
INSERT INTO Tag (name) VALUES ('jQuery'), ('php'), ('asp.net'), ('c#'), ('ruby');
INSERT INTO InterestingTags (user_id, tag_id) VALUES (1,1), (3,1), (3,2), (3,3), (3,4);
/* Select all the users and what tags they are interested in */
SELECT u.name, t.name FROM User u
LEFT JOIN InterestingTags it ON it.user_id = u.id
LEFT JOIN Tag t ON t.id = it.tag_id;
/* Select all tag ids that are interesting to user 3 ("generalist") */
SELECT tag_id FROM InterestingTags WHERE user_id = 3;
/*
Now let's introduce a questions table.
For simplicity, let's say a question can only have one tag.
There's really a many-to-many relationship here, too, as with user and tag
*/
CREATE TABLE Question (id INT NOT NULL AUTO_INCREMENT, title VARCHAR(50) NOT NULL, tag_id INT NOT NULL REFERENCES Tag(id), PRIMARY KEY(id));
/* Insert some questions */
INSERT INTO Question (title, tag_id) VALUES
('generating random numbers in php', 2), /*php question*/
('hiding divs in jQuery', 1), /*jQuery question*/
('how do i add numbers with jQuery', 1), /*jQuery question 2*/
('asp.net help', 3), /*asp.net question */
('c# question', 4), /*c# question */
('ruby question', 5); /*ruby question */
/* select all questions and what users are interested in them */
SELECT q.title, u.name FROM Question q
LEFT JOIN InterestingTags it ON it.tag_id = q.tag_id
LEFT JOIN User u ON u.id = it.user_id;
/* select all questions a user will be interested in. Here the user is jQueryFreak with id = 1 */
SELECT q.id, q.title FROM Question q
LEFT JOIN InterestingTags it ON it.tag_id = q.tag_id
LEFT JOIN User u ON u.id = it.user_id
WHERE u.id = 1;
/* Select all questions and indicate whether or not jQueryFreak (with id = 1) is interested in each one */
/* TODO: make SO question about how to do this as efficient as possible :) */
SELECT q.id, q.title,
(SELECT COUNT(*) FROM InterestingTags it
WHERE it.tag_id = q.tag_id AND it.user_id = 1)
AS is_interested
FROM Question q;
/* Let's add a many-to-many relationship between questions and tags.
Questions can now have many tags
*/
ALTER TABLE Question DROP COLUMN tag_id;
CREATE TABLE Question_Tag (
question_id INT NOT NULL REFERENCES Question (id),
tag_id INT NOT NULL REFERENCES Tag (id),
PRIMARY KEY (question_id, tag_id)
);
/* Insert relationships between questions and tags */
INSERT INTO Question_Tag VALUES
/* First the tags as in the above examples */
(1,2), (2,1), (3,1),(4,3),(5,4),(6,5),
/* And some more. ASP.NET question is also tagged C#
and php question is tagged jQuery */
(1,1), (4,4);
/* select all questions and what users are interested in them
(Some combinations will show up multiple times. This duplication is removed in the
two following queries but I didn't find a solution for it here)*/
SELECT q.title, u.name FROM Question q
LEFT JOIN Question_Tag qt ON qt.question_id = q.id /* <-- new join */
LEFT JOIN InterestingTags it ON it.tag_id = qt.tag_id
LEFT JOIN User u ON u.id = it.user_id;
/* select all questions a user will be interested in. Here the user is jQueryFreak with id = 1 */
SELECT q.id, q.title FROM Question q
LEFT JOIN Question_Tag qt ON qt.question_id = q.id /* <-- new join */
LEFT JOIN InterestingTags it ON it.tag_id = qt.tag_id
LEFT JOIN User u ON u.id = it.user_id
WHERE u.id = 1
GROUP BY q.id; /* prevent duplication of a question in the result list */
/* Select all questions and indicate whether or not jQueryFreak (with id = 1) is interested in each one */
/* STILL TODO: make SO question about how to do this as efficient as possible :) */
SELECT q.id, q.title,
(SELECT COUNT(*) FROM InterestingTags it
WHERE it.tag_id = qt.tag_id AND it.user_id = 1)
AS is_interested
FROM Question q
LEFT JOIN Question_Tag qt ON qt.question_id = q.id /* <-- new join */
GROUP BY q.id;
Update: Added php demo.
Remember to change your mysql constants before running the demo
Ne yapar DB iki sorgu çalışıyor:
- Bir bütün soru ve etiketleri soran
- Kullanıcı etiketleri ne soran tek ilgilendiği
"Mark" kendi etiketleri ile bir soru için, ait olduğu her etiket için bir class ekler - örneğin jQuery (jQuery kimliği vardır 1 burada) ve php (ID ile 2) sınıfları {[olacak etiketlenmiş bir soru (5)]} ve tagged-2.
Şimdi, diğer sorgu ile birleştirerek ilginç etiketleri getiriliyor, sadece ilginç etiketleri karşılık sınıfları olan soruları seçmek ve bunları stil var. Örneğin, ID 1 ve 3, aşağıdaki jQuery kodu olacaktır ile etiketleri ilgileniyorsanız $('.tagged-1, .tagged-3').addClass('interesting-tag');
<?php
const mysql_host = "localhost";
const mysql_username = "";
const mysql_password = "";
const mysql_database = "INTERESTINGTEST";
const user_id = 1; //what user is viewing the page?
class Question {
public $id;
public $title;
public $tags;
function __construct($id,$title) {
$this->id = $id;
$this->title = $title;
$this->tags = array();
}
}
class Tag {
public $id;
public $name;
function __construct($id,$name) {
$this->id = $id;
$this->name = $name;
}
}
/**************************
Getting info from database
****************************/
mysql_connect(mysql_host,mysql_username,mysql_password);
mysql_select_db(mysql_database);
//Fetch interesting tags
$result = mysql_query("SELECT tag_id FROM InterestingTags WHERE user_id = " . user_id);
$interesting_tags = array();
while($row = mysql_fetch_array($result))
{
$interesting_tags[] = $row['tag_id'];
}
//Fetch all questions and their tags
$query_select_questions =
'SELECT q.id AS q_id, q.title AS q_title, t.id AS t_id, t.name AS t_name FROM Question q
LEFT JOIN Question_Tag qt ON qt.question_id = q.id
LEFT JOIN Tag t ON t.id = qt.tag_id';
$result = mysql_query($query_select_questions);
$questions = array();
while($row = mysql_fetch_array($result))
{
$q_id = $row['q_id'];
$q_title = $row['q_title'];
$t_id = $row['t_id'];
$t_name = $row['t_name'];
if (!array_key_exists($q_id, $questions))
$questions[$q_id] = new Question($q_id, $q_title);
$questions[$q_id]->tags[] = new Tag($t_id, $t_name);
}
mysql_close();
/**************************
Write document
****************************/
?>
<style>
.question { padding:0px 5px 5px 5px; border:1px solid gray; margin-bottom: 10px; width:400px }
.interesting-tag { background-color: #FFEFC6 }
</style>
<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.4.2/jquery.js"></script>
<script>
var interesting_tags = [ <?php echo implode($interesting_tags,',') ?> ];
var tagclass_prefix = ".tagged-";
var tags_selector = tagclass_prefix + interesting_tags.join(", " + tagclass_prefix);
$(function() {
$(tags_selector).addClass("interesting-tag");
});
</script>
<?php
foreach ($questions as $q) {
$tagsIDs = array();
$tagNames = array();
foreach ($q->tags as $tag) {
$tagsIDs[] = $tag->id;
$tagNames[] = $tag->name;
}
$classValue = "tagged-" . implode($tagsIDs," tagged-");
$tagNames = implode($tagNames, ", ");
?>
<div id="question-<?php echo $q->id ?>" class="question <?php echo $classValue ?>">
<h3><?php echo $q->title ?></h3>
Tagged with <strong><?php echo $tagNames ?></strong>
</div>
<?php
}
?>
