如何在Stack Overflow上实现类似"有趣的标签"功能？

Question

用赏金检查我的另一个问题: 在表格中查找类似的数字模式

我正在尝试实现一个有趣的标签功能.作为参考,这是它在SO上的工作方式:

1. 我将感兴趣的标签添加到"有趣"列表中(如php,mysql,jquery等).
2. 然后,如果显示的任何问题在我的列表中有一些标记,则会使背景变为橙色.

我理解如何使用jQuery来做到这一点(有相关的问题),但无法弄清楚如何使用MySQL实现后端部分!

所以这是我的问题:它是如何完成的？我想它的工作方式如下:

• 每个成员在mysql中都有一行,我们称之为"interested_tags".
• 在我通过输入写入并提交我的标签后,它将被写入"interested_tags"行.

• 然后,主页面有一个显示所有答案的查询,它总是使用strpos检查带有我的标签的问题标签,如下所示:

  if(strpos($question_tags, $my_tags) === true) {
     //and here will be made background orange
  }
  

我在想是对还是有办法做到这一点？

编辑:那么,你能告诉我一个例子,或者给我一些如何用多对多关系实现这个的技巧吗？谢谢.

Answer 1

正如其他答案中所提到的,用户和标签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InterestingTags表是连接用户对哪些标签感兴趣的表.

/* 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.
在运行演示之前,请记住更改mysql常量

这样做是向DB运行两个查询:

• 一个问所有问题及其标签
• 一个询问用户感兴趣的标签.

要用标签"标记"一个问题,它会class为它所属的每个标签添加一个- 例如标记有jQuery(其中jQuery具有ID 1)和php(带有ID 2)的问题将具有类tagged-1和tagged-2.

现在将此与其他查询相结合,获取有趣的标记,您只需选择具有与有趣标记对应的类的问题并对其进行样式化.举例来说,如果你有兴趣与ID标签1和3,这将是下面的jQuery代码$('.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
    }
?>

Answer 2

每个成员在mysql中都有一行,我们称之为"interested_tags".

更有可能的是,还有一个表格代表用户和标签之间的多对多关系.使用另一个表格将标签与问题相关联.

然后,您只需要一个查询(或更可能是存储过程),将用户的标签与问题的标签进行比较,并返回布尔值true或false.