36

从我表中的两列中，我想获得这些列中值的统一计数。例如，两列是：

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：SELECT type, count(*) from reports group by type

我得到：

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

我试图得到这样的东西：（最右边的一列，我的类型组合在一起，多列，每个地方的计数值）我得到：

|   type        |   home    |   school  |   work    |   cafe    |   friends |   mall    |  
-----------------------------------------------------------------------------------------  
|   one         |   2       |           |   2       |           |           |           |  
|   two         |           |   1       |           |   1       |           |           |  
|   three       |           |           |   2       |           |           |           |  
|   four        |           |           |           |   1       |           |           |  
|   five        |           |           |           |   1       |   1       |           |  
|   six         |           |           |           |           |           |   1       |

这将是像上面这样对每个地方运行计数的结果：

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

这对postgresql有可能吗？

提前致谢。

4

3 回答 3

67

你可以case在这种情况下使用 -

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

它应该可以解决您的问题

@ST Mohammed，要获得这种类型，我们可以简单地在外部查询中使用usingaftergroup或condition，如下所示 -where

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
于 2013-06-12T22:37:24.150 回答
27

praktik garg 的回答是正确的，没有必要使用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，因为布尔值在满足条件时被强制true转换。1

于 2013-06-13T12:01:02.283 回答
17

您也可以使用过滤器子句：

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
于 2017-07-13T14:57:51.013 回答