Question

从我表中的两列开始，我希望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类型，来自报告的count（*）按类别分组

我明白了：

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

我试图得到这样的东西:(一个最右边的列，我的类型组合在一起，多个列与每个地方的计数值）我明白了：

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

这将是对每个地方运行如上所述的计数的结果：

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

这可以用postgresql吗？

提前致谢。

Answer 1

在这种情况下你可以使用case -

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

它应该解决你的问题

@S T Mohammed，要获得此类型，我们只需在外部查询中using或group条件后使用where，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

Answer 2

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，因为当满足条件时，布尔值true将被转换为1。

Answer 3

您也可以使用filter子句：

$role->perms()->sync([]);

Postgresql一个表的多个计数

3 个答案: