0

例如，我有两张桌子

用户

----------------------------------
id  | name | blah
----------------------------------
1   | Samy | stackoverflow
2   | jhon | some thing
----------------------------------

技能

---------------------------------------
id | user_id | skill_title | level
---------------------------------------
1  | 1       | php         | good
2  | 1       | css         | excellent
3  | 1       | photoshop   | fair
4  | 2       | php         | good
---------------------------------------

我像这样运行查询

SELECT * FROM users 
INNER JOIN skills ON users.id = skills.user_id
WHERE ($skill_title[0] LIKE 'skills_title' AND 
       $skill_title[1] LIKE 'skills_title')

其中 $skill_title 是一个数组我需要的是选择具有所有这些技能的用户，即 PHP，CSS，如果我像上面那样查询它永远不会带来数据，因为它将每条记录与所有数组元素进行比较，如果我用或替换它会起作用，但不会给用户带来所有技能

有任何想法吗 ？

4

3 回答 3

1

@joeshmo 的答案可能是您最好的解决方案，但也值得一试。如果您有大量用户，它可能会更快，但很少有人拥有您正在寻找的技能：

SELECT * FROM users
WHERE
    id IN (SELECT user_id FROM skills WHERE skill_title = '$skills[0]')
    AND id IN (SELECT user_id FROM skills WHERE skill_title = '$skills[1]')
    ...

您的 PHP 脚本可能如下所示：

$skills = array("php", "css", ... ); // replace ... with the skills you are looking for
$whereClause = array();
for ($i=0, $count=count($skills); $i < $count; $i++)
    $whereClause[] = "id IN (SELECT user_id FROM skills WHERE skill_title = '{$skills[$i]}')";

$query = "SELECT * FROM users". (count($whereClause) > 0 ? " WHERE ". implode(" AND ", $whereClause) : "");

INNER JOIN正如@joeshmo 建议的那样， 另一种可能的解决方案是使用多个s。通过执行以下操作，您可能可以使其更小更清洁：

SELECT *
FROM
    users u
    INNER JOIN skills s1 ON u.id = s1.user_id AND s1.skill_title = '$skills[0]'
    INNER JOIN skills s2 ON u.id = s2.user_id AND s2.skill_title = '$skills[1]'
    ...

所以你的 PHP 脚本可能看起来像这样：

$skills = array("php", "css", ... );
$joins = array();
for ($i=0, $count=count($skills); $i < $count; $i++)
    $joins[] = "INNER JOIN skills s{$i} ON u.id = s{$i}.user_id AND s{$i}.skill_title = '{$skills[$i]}'";

$query = "SELECT * FROM users u ". implode(" ", $joins);

我会尝试这两种解决方案，看看哪一种对您的数据集表现更好。

于 2012-05-15T13:26:25.943 回答
0

我想你想为你正在寻找的每一项技能再次加入它。

SELECT * FROM users 
INNER JOIN skills as first_skill ON users.id = first_skill.user_id
INNER JOIN skills as second_skill ON users.id = second_skill.user_id
WHERE ($skill_title[0] LIKE first_skill.skill_title AND 
       $skill_title[1] LIKE second_skill.skill_title)

您的查询的问题在于它只加入了技能表一次。您的查询是这样工作的：

-----------------------------------------------------
id  | name | blah           | skill_title | level
-----------------------------------------------------
1   | Samy | stackoverflow  | php         | good
1   | Samy | stackoverflow  | css         | excellent
1   | Samy | stackoverflow  | photoshop   | fair
2   | jhon | some thing     | php         | good
-----------------------------------------------------

没有行同时具有 css 和 php。我的查询是这样工作的：

----------------------------------------------------------------------------------------------------------------
id  | name | blah           | first_skill.skill_title | first_skill.level | second_skill.skill_title | second_skill.level |
----------------------------------------------------------------------------------------------------------------
1   | Samy | stackoverflow  | php                     | good              | php                      | good               |
1   | Samy | stackoverflow  | php                     | good              | css                      | excellent          |
1   | Samy | stackoverflow  | php                     | good              | photoshop                | fair               |
... (there would be nine for Samy)
2   | jhon | some thing     | php                     | good              | php                      
----------------------------------------------------------------------------------------------------------------

第二行同时具有php和css。那就是匹配的那个。

对于多达十个技能，这将是乏味的。您可以改为执行多个查询。

于 2012-05-15T13:18:20.993 回答
0 
$skills = Array("php","css", ..... );
$query = "  select u.id, u.name from users u left join skills s on u.id=s.user_id "
        ."  where s.skill_title in (\"".implode($skills,"\",\"")."\") "
        ."  group by s.user_id having count(s.id) = ".count($skills);
于 2012-05-15T13:26:02.103 回答