如其他答案中所述，用户和标记之间很可能存在多对多关系，表示为自己的表。我做了一个简化案例的SQL演示。该表是连接哪些用户对哪些标签感兴趣的表。InterestingTags
/* 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;
更新：添加了 php 演示。
请记住在运行演示之前更改 mysql 常量
这样做是对数据库运行两个查询：
- 一个询问所有问题及其标签
- 一个询问用户感兴趣的标签。
要用它的标签“标记”一个问题，它为它所属的每个标签添加一个 - 例如，一个标记的问题（其中jQuery有ID）和（带有ID）将有类和。classjQuery1php2tagged-1tagged-2
现在，将此查询与其他查询相结合，获取有趣的标签，您只需要选择具有与有趣标签相对应的类的问题并设置它们的样式即可。例如，如果您对带有ID和的标签感兴趣，它将是以下jQuery代码13$('.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
}
?>
