Question

我有一个表：用户

表字段：

user_id PK INT AI

user_name VARCHAR (45)

parent_user_id FK INT

Rows:
user_id user_name parent_user_id
1       dave            0
2       john            1
3       pritesh         1
4       larry           3
5       abha            4

请创建一个查询，告诉我任何两个用户之间的步数差异。

示例：

user_names: dave and abha
steps difference: 3

说明： dave是pritesh的父母（1步）pritesh是拉里的父母（2步）拉里是abha的父母（3步）。

我已经尝试过以下查询：

SELECT (CASE
         WHEN (count(T2.user_id) - 1) = -1
             THEN 0
             ELSE (count(T2.user_id) - 1)
   END) as steps
FROM
(
SELECT
    @r AS _id,
    (
    SELECT
        @r := parent_user_id
    FROM
        USER
    WHERE
        user_id = _id
) AS parent_user_id,
@l := @l + 1 AS lvl
FROM
(
SELECT
@r := (select MAX(user_id) from USER),
@l := (select MIN(user_id) from USER)
) vars,
USER h
WHERE
@r <> 0
) T1
JOIN
USER T2
ON
T1._id = T2.user_id
WHERE
   (CASE
   WHEN (SELECT user_id FROM USER WHERE user_name= _name1) < (SELECT user_id FROM USER WHERE user_name= _name2)
        THEN (T2.user_id BETWEEN (SELECT user_id FROM USER WHERE user_name= _name1) AND (SELECT user_id FROM USER WHERE user_name= _name2))

   ELSE (T2.user_id BETWEEN (SELECT user_id FROM USER WHERE user_name= _name2) AND (SELECT user_id FROM USER WHERE user_name= _name1))

   END)
ORDER BY
T1.lvl
DESC

Answer 1

宾果@maulik

我找到了你的解决方案，可以找到2个节点之间的一些步骤。我在您的查询中做了一些更改＆amp;让它更容易。

select  count(DISTINCT(parent_user_id)) -1 as Step_Difference
  from    (select * from USER
          order by parent_user_id, user_id) products_sorted,
         (select @pv := 0) initialisation
  where   find_in_set(parent_user_id, @pv)
  and     length(@pv := concat(@pv, ',', user_id))
  AND
  (CASE
      WHEN (SELECT user_id FROM USER WHERE user_name= _name1) < (SELECT user_id FROM USER WHERE user_name= _name2) AND  (SELECT COUNT(user_id) FROM USER WHERE(SELECT user_id FROM USER WHERE user_name = _name1) IN( SELECT parent_user_id FROM USER) ) > 0
          THEN (user_id BETWEEN (SELECT user_id FROM USER WHERE user_name= _name1) AND (SELECT user_id FROM USER WHERE user_name= _name2))

      WHEN (SELECT user_id FROM USER WHERE user_name= _name2) < (SELECT user_id FROM USER WHERE user_name= _name1) AND  (SELECT COUNT(user_id) FROM USER WHERE(SELECT user_id FROM USER WHERE user_name = _name2) IN( SELECT parent_user_id FROM USER) ) > 0
            THEN (user_id BETWEEN (SELECT user_id FROM USER WHERE user_name= _name2) AND (SELECT user_id FROM USER WHERE user_name= _name1))

      ELSE (user_id = 1)

  END)

我怎样才能在mysql中找到父子关系

1 个答案: