首页 / POSTGRCSQL / Postgresql 特性 CTEs (with)

Postgresql 特性 CTEs (with)

内容导读

互联网集市收集整理的这篇技术教程文章主要介绍了Postgresql 特性 CTEs (with)，小编现在分享给大家，供广大互联网技能从业者学习和参考。文章包含6978字，纯文字阅读大概需要10分钟。

内容图文


        CTEs（Common Table Expressions），也就是通用表表达式，你有可能称做它为WITH 语句。和数据库中视图一样，它的主要好处就是，它允许你在当前事务中创建临时表。你可以大量使用它，因为它允许你思路清晰的构建模块，别人很容易就理解你在做什么。        WITH语句作为一个辅助语句依附于主语句，WITH语句和主语句都可以是SELECT，INSERT，UPDATE，DELETE中的任何一种语句。
        CTEs的优势在可读性上，其性能通常不如经过精简优化过的SQL语句性能高。大多数差距小于一倍差距。


让我们select举个简单的例子
WITH users_tasks AS (
  SELECT 
         users.email,
         array_agg(tasks.name) as task_list,
         projects.title
  FROM
       users,
       tasks,
       project
  WHERE
        users.id = tasks.user_id
        projects.title = tasks.project_id
  GROUP BY
           users.email,
           projects.title
)
16 1
WITH users_tasks AS (
2
  SELECT
3
         users.email,
4
         array_agg(tasks.name) as task_list,
5
         projects.title
6
  FROM
7
       users,
8
       tasks,
9
       project
10
  WHERE
11
        users.id = tasks.user_id
12
        projects.title = tasks.project_id
13
  GROUP BY
14
           users.email,
15
           projects.title
16
)


通过这样定义临时表users_tasks，我就可以在后面加上对users_tasks基本查询语句，像：

SELECT *
FROM users_tasks;
2 1
SELECT *
2
FROM users_tasks;

有趣的是你可以将它们连在一起。当我知道分配给每个用户的任务量时，也许我想知道在一个指定的任务上，谁因为对这个任务负责超过了50%而因此造成瓶颈。为了简化，我们可以使用多种方式，先计算每个任务的总量，然后是每人针对每个任务的负责总量。
total_tasks_per_project AS (
  SELECT 
         project_id,
         count(*) as task_count
  FROM tasks
  GROUP BY project_id
),
 
tasks_per_project_per_user AS (
  SELECT 
         user_id,
         project_id,
         count(*) as task_count
  FROM tasks
  GROUP BY user_id, project_id
),
16 1
total_tasks_per_project AS (
2
  SELECT
3
         project_id,
4
         count(*) as task_count
5
  FROM tasks
6
  GROUP BY project_id
7
),
8
9
tasks_per_project_per_user AS (
10
  SELECT
11
         user_id,
12
         project_id,
13
         count(*) as task_count
14
  FROM tasks
15
  GROUP BY user_id, project_id
16
),

现在我们将组合一下然后发现超过50%的用户
overloaded_users AS (
  SELECT tasks_per_project_per_user.user_id
  FROM tasks_per_project_per_user,
       total_tasks_per_project
  WHERE tasks_per_project_per_user.task_count > (total_tasks_per_project / 2)
)
6 1
overloaded_users AS (
2
  SELECT tasks_per_project_per_user.user_id
3
  FROM tasks_per_project_per_user,
4
       total_tasks_per_project
5
  WHERE tasks_per_project_per_user.task_count > (total_tasks_per_project / 2)
6
)


最终目标，我想获得超负荷工作这的用户和任务的逗号分隔列表。我们只要简单地对overloaded_users和 users_tasks的初始列表进行join操作。放在一起可能有点长，但是可读性强。作为额外帮助，我又在每一层加了注释。
--- Query highlights users that have over 50% of tasks on a given project
--- Gives comma separated list of their tasks and the project

--- Initial query to grab project title and tasks per user
WITH users_tasks AS (
  SELECT 
         users.id as user_id,
         users.email,
         array_agg(tasks.name) as task_list,
         projects.title
  FROM
       users,
       tasks,
       project
  WHERE
        users.id = tasks.user_id
        projects.title = tasks.project_id
  GROUP BY
           users.email,
           projects.title
),
 
--- Calculates the total tasks per each project
total_tasks_per_project AS (
  SELECT 
         project_id,
         count(*) as task_count
  FROM tasks
  GROUP BY project_id
),
 
--- Calculates the projects per each user
tasks_per_project_per_user AS (
  SELECT 
         user_id,
         project_id,
         count(*) as task_count
  FROM tasks
  GROUP BY user_id, project_id
),
 
--- Gets user ids that have over 50% of tasks assigned
overloaded_users AS (
  SELECT tasks_per_project_per_user.user_id
  FROM tasks_per_project_per_user,
       total_tasks_per_project
  WHERE tasks_per_project_per_user.task_count > (total_tasks_per_project / 2)
)
 
SELECT 
       email,
       task_list,
       title
FROM 
     users_tasks,
     overloaded_users
WHERE
      users_tasks.user_id = overloaded_users.user_id
58 1
--- Query highlights users that have over 50% of tasks on a given project
2
--- Gives comma separated list of their tasks and the project
3
4
--- Initial query to grab project title and tasks per user
5
WITH users_tasks AS (
6
  SELECT
7
         users.id as user_id,
8
         users.email,
9
         array_agg(tasks.name) as task_list,
10
         projects.title
11
  FROM
12
       users,
13
       tasks,
14
       project
15
  WHERE
16
        users.id = tasks.user_id
17
        projects.title = tasks.project_id
18
  GROUP BY
19
           users.email,
20
           projects.title
21
),
22
23
--- Calculates the total tasks per each project
24
total_tasks_per_project AS (
25
  SELECT
26
         project_id,
27
         count(*) as task_count
28
  FROM tasks
29
  GROUP BY project_id
30
),
31
32
--- Calculates the projects per each user
33
tasks_per_project_per_user AS (
34
  SELECT
35
         user_id,
36
         project_id,
37
         count(*) as task_count
38
  FROM tasks
39
  GROUP BY user_id, project_id
40
),
41
42
--- Gets user ids that have over 50% of tasks assigned
43
overloaded_users AS (
44
  SELECT tasks_per_project_per_user.user_id
45
  FROM tasks_per_project_per_user,
46
       total_tasks_per_project
47
  WHERE tasks_per_project_per_user.task_count > (total_tasks_per_project / 2)
48
)
49
50
SELECT
51
       email,
52
       task_list,
53
       title
54
FROM
55
     users_tasks,
56
     overloaded_users
57
WHERE
58
      users_tasks.user_id = overloaded_users.user_id

来个delete的例子：
本例通过WITH中的DELETE语句从products表中删除了一个月的数据，并通过RETURNING子句将删除的数据集赋给moved_rows这一CTE，最后在主语句中通过INSERT将删除的商品插入products_log中。
WITH moved_rows AS (
  DELETE FROM products
  WHERE
    "date" >= ‘2010-10-01‘
  AND "date" < ‘2010-11-01‘
  RETURNING *
)
INSERT INTO products_log
SELECT * FROM moved_rows;
9 1
WITH moved_rows AS (
2
  DELETE FROM products
3
  WHERE
4
    "date" >= ‘2010-10-01‘
5
  AND "date" < ‘2010-11-01‘
6
  RETURNING *
7
)
8
INSERT INTO products_log
9
SELECT * FROM moved_rows;

如果WITH里面使用的不是SELECT语句，并且没有通过RETURNING子句返回结果集，则主查询中不可以引用该CTE，但主查询和WITH语句仍然可以继续执行。这种情况可以实现将多个不相关的语句放在一个SQL语句里，实现了在不显式使用事务的情况下保证WITH语句和主语句的事务性。


WITH使用注意事项【个人感觉有点类似线程不安全】

        WITH中的数据修改语句会被执行一次，并且肯定会完全执行，无论主语句是否读取或者是否读取所有其输出。而WITH中的SELECT语句则只输出主语句中所需要记录数。 WITH中使用多个子句时，这些子句和主语句会并行执行，所以当存在多个修改子语句修改相同的记录时，它们的结果不可预测。 所有的子句所能“看”到的数据集是一样的，所以它们看不到其它语句对目标数据集的影响。这也缓解了多子句执行顺序的不可预测性造成的影响。 如果在一条SQL语句中，更新同一记录多次，只有其中一条会生效，并且很难预测哪一个会生效。 如果在一条SQL语句中，同时更新和删除某条记录，则只有更新会生效。 目前，任何一个被数据修改CTE的表，不允许使用条件规则，和ALSO规则以及INSTEAD规则。


WITH RECURSIVE

        WITH语句还可以通过增加RECURSIVE修饰符来引入它自己，从而实现递归。        WITH RECURSIVE一般用于处理逻辑上层次化或树状结构的数据，典型的使用场景是寻找直接及间接子结点。




Postgresql 特性 CTEs (with)

标签：keyword   cal   简单   osi   事务性   任务   relative   group by   ali   

本文系统来源：https://www.cnblogs.com/churao/p/8494315.html

内容总结

以上是互联网集市为您收集整理的Postgresql 特性 CTEs (with)全部内容，希望文章能够帮你解决Postgresql 特性 CTEs (with)所遇到的程序开发问题。 如果觉得互联网集市技术教程内容还不错，欢迎将互联网集市网站推荐给程序员好友。

内容备注

版权声明：本文内容由互联网用户自发贡献，该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务，不拥有所有权，不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容， 请发送邮件至 gblab@vip.qq.com 举报，一经查实，本站将立刻删除。

内容手机端

扫描二维码推送至手机访问。

本文链接：https://qyyshop.com/info/499850.html

来源：【匿名】