Question

你的帮助将不胜感激...... 如果我有下表和示例数据... myGroupTable

group_Id：user_Id
1：3
1：7
1：100
2：3
2：7
2：100
2：104
4：42
4：98
4：13

我想要一个sql语句，它会...... 返回一个完全具有指定user_Id的group_Id。例如......是否有group_Id具有User_Id的3,7和100 回答：group_id 1。请注意，我不希望它返回group_Id 2，因为它的user_Id为104 ... 亲切的问候J

Answer 1

SELECT
    group_Id,
    SUM(
        IF user_Id = 3 THEN 1
        ELSEIF user_Id = 7 THEN 2
        ELSEIF user_Id = 100 THEN 4
        ELSE 8
    ) AS bits
FROM myGroupTable
GROUP BY group_Id
HAVING bits=7

这假设你不能为同一个group_Id拥有重复的user_Id，例如，这可能永远不会发生：

group     user
   1         3
   1         3

修改：您可以按以下方式构建查询：

<?php
$ids = array(3, 7, 100);
$power = 2;
$query = "
    SELECT
    group_Id,
    SUM(
        IF user_Id = " .$ids[0]. " THEN 1 ";
foreach ($id in $ids) { 
    $query .= " ELSEIF user_Id = $id THEN " . $power;
    $power = $power * 2;
}
$query .= " ELSE $power
    ) AS bits
    FROM myGroupTable
    GROUP BY group_Id
    HAVING bits = " . ($power - 1);

Answer 2

这是另一种选择：

SELECT group_id, GROUP_CONCAT(user_id ORDER BY user_id) AS user_id_list
FROM group_user
GROUP BY group_id
HAVING user_id_list = '3,7,100'

Answer 3

另一种解决方案：

SELECT group FROM tbl WHERE group_id NOT IN（SELECT DISTINCT group_id FROM tbl WHERE user_id NOT IN（3,7,100））AND user_id IN（3,7,100）;

Answer 4

select group_id,
sum(
  case user_id in(3,7,100)
  when 1 then 1
  else -99
  end
) as must_be_3
from group_user
group by group_id
having must_be_3=3;

MYSQL基于选择选择

4 个答案: