一个表的Postgresql多重计数

从表中的两列中，我希望获得这些列中值的统一计数。 例如，两列为：

表：报告

|   type        |   place   |  
 ----------------------------------------- 
|   one         |   home    |  
|   two         |   school  |  
|   three       |   work    |  
|   four        |   cafe    |  
|   five        |   friends |  
|   six         |   mall    |  
|   one         |   work    |  
|   one         |   work    |  
|   three       |   work    |  
|   two         |   cafe    |  
|   five        |   cafe    |  
|   one         |   home    |  

如果我这样做： 从报告中选择类型、计数（*） 按类型分组

我得到：

|   type        |   count   |  
-----------------------------  
|   one         |   4       |  
|   two         |   2       |  
|   three       |   2       |  
|   four        |   1       |  
|   five        |   2       |  
|   six         |   1       | 

我试图得到这样的结果：（最右边的一列和我的类型分组在一起，多个列和每个位置的计数值） 我得到：

这将是对每个这样的地方进行上述计数的结果：

SELECT type, count(*) from reports where place  = 'home'
group by type
SELECT type, count(*) from reports where place  = 'school'
group by type
SELECT type, count(*) from reports where place  = 'work'
group by type
SELECT type, count(*) from reports where place  = 'cafe'
group by type
SELECT type, count(*) from reports where place  = 'friends'
group by type
SELECT type, count(*) from reports where place  = 'mall'
group by type

postgresql是否可以实现这一点

---

提前感谢。

在这种情况下，您可以使用

案例

-

SELECT type, 
       sum(case when place  = 'home' then 1 else 0 end) as Home,
       sum(case when  place  = 'school' then 1 else 0 end) as school,
       sum(case when  place  = 'work' then 1 else 0 end) as work,
       sum(case when  place  = 'cafe' then 1 else 0 end) as cafe,
       sum(case when  place  = 'friends' then 1 else 0 end) as friends,
       sum(case when  place  = 'mall' then 1 else 0 end) as mall
  from reports
 group by type

它应该能解决你的问题

@S T穆罕默德， 要获得这样的类型，我们只需在外部查询中使用

use

after

group

或

where

条件，如下所示-

select type, Home, school, work, cafe, friends, mall from (
SELECT type, 
       sum(case when place  = 'home' then 1 else 0 end) as Home,
       sum(case when  place  = 'school' then 1 else 0 end) as school,
       sum(case when  place  = 'work' then 1 else 0 end) as work,
       sum(case when  place  = 'cafe' then 1 else 0 end) as cafe,
       sum(case when  place  = 'friends' then 1 else 0 end) as friends,
       sum(case when  place  = 'mall' then 1 else 0 end) as mall
  from reports
 group by type
 )
 where home >0 and School >0 and Work >0 and cafe>0 and friends>0 and mall>0

---

praktik garg的回答是正确的，不必使用

else 0

：

SELECT type, 
       sum(case when place  = 'home' then 1 end) as home,
       sum(case when  place  = 'school' then 1 end) as school,
       sum(case when  place  = 'work' then 1 end) as work,
       sum(case when  place  = 'cafe' then 1 end) as cafe,
       sum(case when  place  = 'friends' then 1 end) as friends,
       sum(case when  place  = 'mall' then 1 end) as mall
FROM reports
GROUP BY type

您还可以使用以下更短的语法：

SELECT type, 
       sum((place  = 'home')::int) as home,
       sum((place  = 'school')::int) as school,
       sum((place  = 'work' )::int) as work,
       sum((place  = 'cafe' )::int) as cafe,
       sum((place  = 'friends')::int) as friends,
       sum((place  = 'mall')::int) as mall
FROM reports
GROUP BY type

---

这将起作用，因为当满足条件时，布尔值

true

被转换为

1

。

您也可以使用filter子句：

SELECT
  type,
  sum(1) FILTER (WHERE place = 'home') AS home,
  sum(1) FILTER (WHERE place = 'school') AS school,
  sum(1) FILTER (WHERE place = 'work') AS work,
  sum(1) FILTER (WHERE place = 'cafe') AS cafe,
  sum(1) FILTER (WHERE place = 'friends') AS friends,
  sum(1) FILTER (WHERE place = 'mall') AS mall
FROM
  reports
GROUP BY 
  type

---

要获得空（null）列而不是零，请将总和包装在

nullif

calls中。对失败案例使用

count

和

null

而不是

0

（完全省略

ELSE 0

）也可以，您不必更改为

sum

。sql语句中的最后一个逗号（就在“from”关键字之前）会导致错误。如何获取家庭、学校、工作、咖啡馆、朋友的记录，mall values>0对此延迟回复表示歉意。通过在查询中做一些小的更改，您可以得到您想要的输出。例如（‘home’、‘school’…）我有可能只在运行时知道的位置查找。有什么样的sql模板吗？

SELECT
  type,
  sum(1) FILTER (WHERE place = 'home') AS home,
  sum(1) FILTER (WHERE place = 'school') AS school,
  sum(1) FILTER (WHERE place = 'work') AS work,
  sum(1) FILTER (WHERE place = 'cafe') AS cafe,
  sum(1) FILTER (WHERE place = 'friends') AS friends,
  sum(1) FILTER (WHERE place = 'mall') AS mall
FROM
  reports
GROUP BY 
  type