递归查询postgresql

Question

我在使用 sql 进行递归查询时遇到问题。我想知道解决这个问题的最佳方法是什么。

我有以下型号

这里应用程序基本上是有版本的，安装在不同的环境中。需要注意的是，每个版本都有一个以前的版本，因此可以应用类似于 hotfix 的模型。要知道环境中的工单，需要搜索最新版本的环境并进行递归

我的目标是构建一个 table 如下：

门票 | arrival_date_to_enviroment | arrival_version_to_enviroment |环境

我想我必须进行递归查询，才能知道哪些票在哪个版本中。为此，我认为第一步是了解版本包含的版本我可以通过以下方式提出一些问题来解决这部分问题：

  WITH RECURSIVE version_recursive AS (
    select v.name, v.version_from, v.application  from versions v 
    where v.name='21.1204.0'
        UNION
    select v.name, v.version_from, v.application  from versions v 
    JOIN version_recursive s ON s.version_from = v.name AND s.application = v.application
        )
 SELECT *
   FROM version_recursive;

下一步是在环境中找到最新的部署并使用票证加入。

但我的主要问题之一是我不知道如何使用 sql

完成 arrival_version_to_enviroment

非常感谢任何帮助和建议。我需要使用 sql 来完成此操作，无需编程逻辑

Answer 1

如果我理解你到达的意思，这样的事情可能会做到：

遍历版本（递归），但保留每个相关先前版本的基础版本。
加入各种需要的细节

我在结果中包含了比要求的更多的细节，并且可能建议了更多。例如，了解引入票证的初始环境很有用，但此部署的当前环境也是如此。我决定只包括引入的初始环境。添加更多细节很简单。

SQL:

WITH RECURSIVE versionx (base, name, version_from, application, lv) AS (
         SELECT name, name, version_from, application, 1 FROM versions
          UNION  ALL
         SELECT v0.base, v1.name, v1.version_from, v1.application, lv+1
           FROM versions AS v1
           JOIN versionx AS v0
             ON v0.version_from = v1.name
            AND lv < 20                     -- Temporary guard against bad data. Remove when ready.
     )
SELECT t.id    AS ticket_id         -- The ticket
     , d.date  AS arrival_date      -- Initial date ticket was introduced.
     , x.name  AS arrival_version   -- Initial version ticket was introduced.
     , x.base  AS version           -- Current version.
     , d0.date AS cur_date          -- Date of deployment of current version.
     , e.name AS env_name           -- Initial deployment environment
  FROM versionx AS x
  JOIN tickets  AS t
    ON t.version = x.name
  JOIN deployments   AS d
    ON d.version = x.name
  JOIN environments  AS e
    ON e.id = d.environment_id
  JOIN deployments   AS d0
    ON d0.version = x.base
;

结果，基于给定的测试数据：

+-----------+--------------+-----------------+---------+------------+----------+
| ticket_id | arrival_date | arrival_version | version | cur_date   | env_name |
+-----------+--------------+-----------------+---------+------------+----------+
|         1 | 2021-12-07   | ver01           | ver01   | 2021-12-07 | env01    |
|         1 | 2021-12-07   | ver01           | ver02   | 2021-12-08 | env01    |
|         2 | 2021-12-08   | ver02           | ver02   | 2021-12-08 | env02    |
+-----------+--------------+-----------------+---------+------------+----------+

测试数据：

SELECT * FROM versions;

+-------+--------------+-------------+
| name  | version_from | application |
+-------+--------------+-------------+
| ver01 | NULL         | app01       |
| ver02 | ver01        | app02       |
+-------+--------------+-------------+

SELECT * FROM deployments;

+------+----------------+------------+---------+
| id   | environment_id | date       | version |
+------+----------------+------------+---------+
|    1 |              1 | 2021-12-07 | ver01   |
|    2 |              2 | 2021-12-08 | ver02   |
+------+----------------+------------+---------+

SELECT * FROM environments;

+------+-------+-------------+-------+
| id   | name  | application | url   |
+------+-------+-------------+-------+
|    1 | env01 | app01       | url01 |
|    2 | env02 | app02       | url02 |
+------+-------+-------------+-------+

SELECT * FROM tickets;

+------+---------+
| id   | version |
+------+---------+
|    1 | ver01   |
|    2 | ver02   |
+------+---------+

Answer 2

不同的例子：

-- creating table 
CREATE TABLE versions (
    "name" text NULL,
    version_from text NULL,
    aqpplication text NULL
);

-- insert sample data for test 
INSERT INTO versions ("name", version_from, aqpplication) VALUES('1.0',  NULL,   'app1');
INSERT INTO versions ("name", version_from, aqpplication) VALUES('1.1',  '1.0',  'app1');
INSERT INTO versions ("name", version_from, aqpplication) VALUES('1.2',  '1.1',  'app1');
INSERT INTO versions ("name", version_from, aqpplication) VALUES('1.3',  '1.21', 'app1');
INSERT INTO versions ("name", version_from, aqpplication) VALUES('5.5',  NULL,   'app2');
INSERT INTO versions ("name", version_from, aqpplication) VALUES('5.6',  '5.5',  'app2');
INSERT INTO versions ("name", version_from, aqpplication) VALUES('5.7',  '5.6',  'app2');
INSERT INTO versions ("name", version_from, aqpplication) VALUES('5.8',  '5.7',  'app2');
INSERT INTO versions ("name", version_from, aqpplication) VALUES('1.4',  '1.3',  'app1');


-- recursive select query 
WITH RECURSIVE version_recursive(name, version_from, application, version_history) AS (
 SELECT vn.*, vn."name"::text
 FROM versions AS vn 
 WHERE vn.version_from is null 
UNION ALL
 SELECT v.*,  
        (v_rec.version_history || '->' || v."name"::TEXT) 
 FROM version_recursive as v_rec, versions AS v 
 WHERE v.version_from = v_rec."name"
)
SELECT * FROM version_recursive ORDER BY "name";


Result: 
+--------+--------------+-------------+--------------------+
| name   | version_from | application | version_history    |
+--------+--------------+-------------+--------------------+
| 1.0    | NULL         | app1        | 1.0                |
| 1.1    | 1.0          | app1        | 1.0->1.1           |
| 1.2    | 1.1          | app1        | 1.0->1.1->1.2      |
| 5.5    | NULL         | app2        | 5.5                |
| 5.6    | 5.5          | app2        | 5.5->5.6           |
| 5.7    | 5.6          | app2        | 5.5->5.6->5.7      |
| 5.8    | 5.7          | app2        | 5.5->5.6->5.7->5.8 |
+--------+--------------+-------------+--------------------+

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