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Mysql 5.7 - 如何执行多个需要前一个子查询结果的子查询？

[英]Mysql 5.7 - how to execute multiple subqueries that need the result of the previous subquery?

 原文  2021-03-10 03:30:55  6   1    mysql/ linux/ subquery/ left-join

问题描述

希望有人可以帮助我解决我认为是子查询问题 - 我无法解决的问题。

我有一个包含用户推荐信息的表，推荐可以是 3 层。 例如。

吉姆提到史蒂夫

瑞安提到吉姆

斯蒂芬妮提到瑞恩

马克提到斯蒂芬妮

作为更大查询的一部分，我试图做的是检索用户“史蒂夫”的推荐人列表所以理想情况下，查询返回“

  • 吉姆
  • 瑞安
  • 斯蒂芬妮
  • 标记

让您了解“user_referrals”表，如果史蒂夫是用户 ID 6211

--

mysql> select * from user_referrals where referee = '6211';
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
| id   | referrer | referee | created                    | modified                   | isActualReferrer | deleted | version |
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
| 4934 |     6155 |    6211 | 2021-02-28 04:22:35.798832 | 2021-02-28 04:22:35.798832 |                1 | NULL    |       1 |
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
1 row in set (0.00 sec)

mysql> select* from user_referrals where referee = '6155';
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
| id   | referrer | referee | created                    | modified                   | isActualReferrer | deleted | version |
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
| 4881 |     5757 |    6155 | 2021-02-25 17:44:42.103963 | 2021-02-25 17:44:42.103963 |                1 | NULL    |       1 |
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
1 row in set (0.00 sec)

mysql> select * from user_referrals where referee = '5757';
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
| id   | referrer | referee | created                    | modified                   | isActualReferrer | deleted | version |
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
| 4589 |       29 |    5757 | 2020-12-30 05:02:23.831882 | 2020-12-30 05:02:23.831882 |                1 | NULL    |       1 |
+------+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
1 row in set (0.00 sec)

mysql> select * from user_referrals where referee = '29';
+----+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
| id | referrer | referee | created                    | modified                   | isActualReferrer | deleted | version |
+----+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
|  7 |       24 |      29 | 2019-02-02 00:22:12.000000 | 2019-02-02 00:22:12.000000 |                1 | NULL    |       1 |
+----+----------+---------+----------------------------+----------------------------+------------------+---------+---------+
1 row in set (0.01 sec)

主查询的子查询部分如下所示

    LEFT JOIN
    
    ( select u1.referrer from user_referrals as u1 ) as u ON (u.id = u.referrer)
    
    LEFT JOIN
    
    ( select u1.referrer from user_referrals as u1 ) as u3 ON (u3.referrer = (
    select u2.referrer from user_referrals as u2 WHERE u.referrer = u2.referee)
    )

我不认为这是处理它的最佳方法，在主查询中使用上述内容，到目前为止我得到的结果如下：

示例查询

select distinct(u.id),u.created,u.first_name,p.amount,u.referrer,u3.referrer from users as u
    LEFT JOIN
    ( Select p1.* FROM paymentledger as p1 LEFT JOIN paymentledger as p2 ON p1.userID = p2.userID AND p1.createdAt > p2.createdAt WHERE p2.userID IS NULL) as p ON (u.id = p.userID)
    LEFT JOIN
    
    ( select u1.referrer from user_referrals as u1 ) as u ON (u.id = u.referrer)
    
    LEFT JOIN
    
    ( select u1.referrer from user_referrals as u1 ) as u3 ON (u3.referrer = (
    select u2.referrer from user_referrals as u2 WHERE u.referrer = u2.referee)
    )
    WHERE u.created LIKE '2021-02-28%';

样品 output

+------+----------------------------+------------------+--------+----------+----------+
| id   | created                    | first_name       | amount | referrer | referrer |
+------+----------------------------+------------------+--------+----------+----------+
| 6202 | 2021-02-28 00:00:12.903938 | Brian            |   NULL |     NULL |     NULL |
| 6204 | 2021-02-28 00:44:52.317175 | Katie            |   NULL |     NULL |     NULL |
| 6205 | 2021-02-28 01:28:14.379496 | Stephanie        | 159.00 |     NULL |     NULL |
| 6206 | 2021-02-28 01:32:45.655260 | Christina        |   NULL |     NULL |     NULL |
| 6207 | 2021-02-28 01:54:22.954110 | R                |   NULL |     NULL |     NULL |
| 6208 | 2021-02-28 02:45:01.010985 | Lillie           |   NULL |     NULL |     NULL |
| 6209 | 2021-02-28 02:53:10.819064 | Deborah          |   NULL |     NULL |     NULL |
| 6210 | 2021-02-28 04:05:24.937563 | Danielle         |   NULL |     NULL |     NULL |
| 6211 | 2021-02-28 04:22:35.103951 | Linda            | 159.00 |     6211 |     6155 |
| 6212 | 2021-02-28 04:32:53.045521 | Kathy            |   NULL |     NULL |     NULL |
| 6213 | 2021-02-28 05:05:17.063022 | malekmuneerah786 |   NULL |     NULL |     NULL |
| 6214 | 2021-02-28 07:47:12.725790 | James            |   NULL |     NULL |     NULL |
| 6215 | 2021-02-28 08:10:01.976589 | Wendy            |   NULL |     NULL |     NULL |
| 6216 | 2021-02-28 08:49:09.520211 | Barbara          |   NULL |     NULL |     NULL |
| 6217 | 2021-02-28 08:53:31.673631 | Tammy            |   NULL |     NULL |     NULL |
| 6218 | 2021-02-28 09:07:59.438093 | San              |   NULL |     NULL |     NULL |
| 6219 | 2021-02-28 09:30:26.966068 | Joan R           |   NULL |     NULL |     NULL |
| 6220 | 2021-02-28 14:58:11.305224 | Moses            |   NULL |     NULL |     NULL |
| 6221 | 2021-02-28 17:26:51.210201 | De Ana           |   NULL |     NULL |     NULL |
| 6222 | 2021-02-28 17:27:25.950142 | Faisal           |   NULL |     NULL |     NULL |
| 6223 | 2021-02-28 19:25:43.374723 | Pauline          |   NULL |     NULL |     NULL |
| 6224 | 2021-02-28 19:55:34.650488 | Jana             | 159.00 |     6224 |     5757 |
| 6225 | 2021-02-28 20:13:23.158743 | Authentically    |   NULL |     NULL |     NULL |
| 6226 | 2021-02-28 20:17:04.178635 | J                |   NULL |     NULL |     NULL |
| 6227 | 2021-02-28 21:49:59.988693 | JACOBINA         |   NULL |     NULL |     NULL |
| 6228 | 2021-02-28 21:59:00.075468 | Jess             |   NULL |     NULL |     NULL |
| 6229 | 2021-02-28 22:55:16.655047 | Junie            |   NULL |     NULL |     NULL |
+------+----------------------------+------------------+--------+----------+----------+

可能只是我对加入感到困惑，希望有人可以在这里提供一些指导。

提前谢谢你。

  • 根据要求 - 添加 user_referrals 的创建表和示例输出/所需的 output。
CREATE TABLE `user_referrals` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `referrer` int(11) DEFAULT NULL,
  `referee` int(11) DEFAULT NULL,
  `created` datetime(6) DEFAULT CURRENT_TIMESTAMP(6),
  `modified` datetime(6) DEFAULT CURRENT_TIMESTAMP(6),
  `isActualReferrer` tinyint(4) NOT NULL DEFAULT '1',
  `deleted` datetime(6) DEFAULT NULL,
  `version` int(11) NOT NULL DEFAULT '1',
  PRIMARY KEY (`id`)
  );

+------+----------------------------+------------------+--------+----------+----------+----------+----------+
| id   | created                    | first_name       | amount | referrer | referrer | referrer | referrer |
+------+----------------------------+------------------+--------+----------+----------+----------+----------+
| 6210 | 2021-02-28 04:05:24.937563 | Danielle         |   NULL |     NULL |     NULL |     NULL |     NULL |
| 6211 | 2021-02-28 04:22:35.103951 | Linda            | 159.00 |     6211 |     6155 |     5757 |     29   |
| 6212 | 2021-02-28 04:32:53.045521 | Kathy            |   NULL |     NULL |     NULL |     NULL |     NULL |
+------+----------------------------+------------------+--------+----------+----------+----------+----------+
27 rows in set (0.04 sec)

这是user_referrals表的demo表，创建代码+插入demo数据

CREATE TABLE `user_referrals_demo` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `referrer` int(11) DEFAULT NULL,
  `referee` int(11) DEFAULT NULL,
  `created` datetime(6) DEFAULT CURRENT_TIMESTAMP(6),
  `modified` datetime(6) DEFAULT CURRENT_TIMESTAMP(6),
  `isActualReferrer` tinyint(4) NOT NULL DEFAULT '1',
  `deleted` datetime(6) DEFAULT NULL,
  `version` int(11) NOT NULL DEFAULT '1',
  PRIMARY KEY (`id`)
  );

INSERT INTO user_referrals_demo(referrer,referee) VALUES ('6155','6211'),('5757','6155'),('29','5757'),('24','29');

output 应该是

用户名 | 推荐人 1 | 推荐人 2 | 推荐人 3 | 推荐人 4 6211 | 6155 | 5757 | 29 | 24

我正在尝试通过记录递归 sql 查询（我认为我需要一个子查询）来递归 go 以提取所需的数据。

1 个解决方案

解决方案1
 0  2021-03-10 10:41:21

mysql> select u3.referrer as referrer3,
    -> u2.referrer as referrer2,
    -> u1.referrer as referrer1
    -> from user_referrals_demo as u1
    ->
    -> left join user_referrals as u2 ON (u2.referee = u1.referrer)
    -> left join user_referrals as u3 ON (u3.referee = u2.referrer)
    -> left join user_referrals as u4 ON (u4.referee = u3.referrer)
    ->
    -> where u1.referee = '6211';

上面的 sql 提供了所需的 output ，通过这段代码，我能够将其用于父查询。

输出

+-----------+-----------+-----------+
| referrer3 | referrer2 | referrer1 |
+-----------+-----------+-----------+
|        29 |      5757 |      6155 |
+-----------+-----------+-----------+

感谢帮助我解决此问题的有关此问题的私人消息。

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