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Question

我在users表中有一个名为skills的列，其类型为VARCHAR，值为implod，由PHP编写。例如1,2,3,4,5

我正在尝试制作作品列表，作品具有相同的技能组合（不同的ID ），每个work都有自己的技能列，其格式与前一个。例如2,6,5,3,7

所以我想将这两个列与查询进行比较，看看作品是否具有用户的全部或部分技能，如果是，则显示它。

正如已发布的社区主题，我发现了FIND_IN_SET或多个LIKE语句的一些解决方案，但实际上它们不适合我的情况。

由于我不是查询粉丝，所以这种比较的最佳或最佳做法是什么？

Answer 1

最佳解决方案是更改架构。不要在一列中存储多个值。用户和技能之间的关系是“多对多”的关系。对于任何多对多关系，您应该使用三个表：

users -- one record for each user
+----+------+
| id | name |
+----+------+
| 1  | Bob  |
| 2  | Joe  |
| 3  | Ken  |
+----+------+

skills -- one record for each possible skill
+----+-------------+
| id | name        |
+----+-------------+
| 1  | PHP         |
| 2  | CSS         |
| 3  | JS          |
| 4  | MySQL       |
+----+-------------+

userSkills -- relational table that maps skills to users
+--------+---------+
| userId | skillId |
+--------+---------+
| 1      | 1       |
| 1      | 2       |
| 1      | 3       |
| 1      | 4       |
| 2      | 2       |
| 2      | 3       |
| 2      | 4       |
| 3      | 1       |
| 3      | 4       |
+--------+---------+

现在，当您想要了解Joe的技能时，您可以执行以下查询：

SELECT
    s.name
FROM
    users u
    INNER JOIN userSkills us
        ON u.id = us.userId
    INNER JOIN skills s
        ON us.skillId = s.id
WHERE
    u.name = 'Joe'

作品和技能之间的关系也是多对多的关系。我们还需要两个表，因为我们已经有了skills表。

works
+----+-----------------------+
| id | name                  |
+----+-----------------------+
| 1  | Client-side developer |
| 2  | Server-side developer |
| 3  | All-around developer  |
+----+-----------------------+

workSkills
+--------+---------+
| workId | skillId |
+--------+---------+
| 1      | 2       |
| 1      | 3       |
| 2      | 1       |
| 2      | 4       |
| 3      | 1       |
| 3      | 2       |
| 3      | 3       |
| 3      | 4       |
+--------+---------+

现在，您可以通过以下查询找到服务器端开发人员所需的所有技能：

SELECT
    s.name
FROM
    works w
    INNER JOIN workSkills ws
        ON w.id = ws.workId
    INNER JOIN skills s
        ON ws.skillId = s.id
WHERE
    w.name = 'Server-side developer'

现在是棘手的部分。获取给定工作的用户列表。以下查询将为您提供具有与指定工作相同技能的用户列表。他们对这项工作的技能越多，他们的得分就越高：

SELECT
    u.name,
    COUNT(s.name) AS score
FROM
    users u
    INNER JOIN userSkills us
        ON u.id = us.userId
    INNER JOIN skills s
        ON us.skillId = s.id
    INNER JOIN workSkills ws
        ON s.id = ws.skillId
    INNER JOIN works w
        ON ws.workId = w.id
  WHERE
      w.name = 'Server-side developer'
  GROUP BY
      u.name
  ORDER BY
      score DESC

输出：

+------+-------+
| name | score |
+------+-------+
| Bob  | 2     | -- Bob has all four skills, but only PHP and MySQL are relevant
| Ken  | 2     | -- Ken has PHP and MySQL skills, same as this work requires
| Joe  | 1     | -- Joe has CSS, JS, and MySQL skills...only MySQL is relevant
+------+-------+

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Answer 2

我建议你不要让技能ids像那样崩溃。保留5个表格。

用户（用户ID，名称）
技能（技能，技能名称）
usersskills（skillid，userid）
作品（workid，name）
workskills（workid，skillid）

您将拥有干净且可扩展的结构，并且易于提取数据。

获取用户技能的SQL是

SELECT skillid, skillname FROM Skills INNER JOIN UsersSkills ON Skills.skillid = UsersSkills.skillid WHERE UserSkills.userid = 3

获取作品技巧的SQL

SELECT skillid, skillname FROM skills INNER JOIN workskills ON skills.skillid = workskills.skillid WHERE workskills.workid = 1

我们假设我们有这两个表

usersskills              worksskills
+--------+---------+     +--------+-------+
|skillid | userid  |     |skillid |workid |
+--------+---------+     +--------+-------+
| 1      | 1       |     | 1      | 3     |
| 2      | 1       |     | 2      | 3     |
| 3      | 1       |     | 3      | 3     |
| 4      | 1       |     | 4      | 3     |
| 1      | 2       |     | 5      | 3     |
| 2      | 2       |     +--------+--------
+--------+---------+

这里我们有2个用户，第一个有4个技能，第二个有2个技能。我们的工作（身份3）需要5种不同的技能。

以下是此项工作的用户资格。

+--------+-------------+
|userid  |scorepercent |
+--------+-------------+
|1       |80           |
|2       |40           |
+--------+-------------+

用户1的工作资格为80％，用户2资格为40％

SQL（简短和干净）

按分数排序的所有用户

select 
    userid,
    (count(skillid) * 100 / (select count(*) from skillswork where workid = 3)) as score
from
    skillsuser
where
    skillid in (select skillid from skillswork where workid = 3)
group by
    userid
order by 
    score desc

如果您不需要百分比，只需要一个简单的SQL

select 
    userid,
    count(skillid) as score
from
    skillsuser
where
    skillid in (select skillid from skillswork where workid = 3)
group by
    userid
order by 
    score desc

单个目标用户

select 
    userid,
    (count(skillid) * 100 / (select count(*) from skillswork where workid = 3)) as score
from
    skillsuser
where
    skillid in (select skillid from skillswork where workid = 3)
    and
    userid = 1

SQLFiddle

结果基于用户技能

2 个答案:

SQLFiddle example