我需要知道如何创建一个交叉表查询

Question

我需要帮助创建下面的结果。我想到了一个SQL数据透视表，但我不知道如何使用它。看了几个例子，不能提出解决方案。任何其他想法如何实现这一点也是受欢迎的。状态列必须动态生成。

有三个表，资产，assettypes，assetstatus

 
Table: assets 
assetid  int 
assettag varchar(25) 
assettype int 
assetstatus int 

Table: assettypes 
id   int 
typename varchar(20) (ex: Desktop, Laptop, Server, etc.) 

Table: assetstatus 
id   int 
statusname varchar(20) (ex: Deployed, Inventory, Shipped, etc.) 

期望的结果：

 
AssetType  Total Deployed Inventory Shipped  ... 
----------------------------------------------------------- 
Desktop   100  75  20   5  ... 
Laptop   75  56  19   1  ... 
Server   60  50  10   0  ... 

一些数据：

 
assets table: 
1,hol1234,1,1 
2,hol1233,1,2 
3,hol3421,2,3 
4,svr1234,3,1 

assettypes table: 
1,Desktop 
2,Laptop 
3,Server 

assetstatus table: 
1,Deployed 
2,Inventory 
3,Shipped 

Answer 1

这种类型的转换被称为支点。你没有指定你正在使用的数据库，所以我将提供SQL Server和MySQL的答案。

---

SQL服务器：如果您正在使用SQL Server 2005+就可以实现PIVOT功能。

如果您有已知数量的值要转换为列，那么您可以硬编码查询。

select typename, total, Deployed, Inventory, shipped 
from 
(
    select count(*) over(partition by t.typename) total, 
    s.statusname, 
    t.typename 
    from assets a 
    inner join assettypes t 
    on a.assettype = t.id 
    inner join assetstatus s 
    on a.assetstatus = s.id 
) d 
pivot 
(
    count(statusname) 
    for statusname in (Deployed, Inventory, shipped) 
) piv; 

请参阅SQL Fiddle with Demo。

但是，如果您有一个未知数的status值，那么您将需要使用动态sql在运行时生成列的列表。

DECLARE @cols AS NVARCHAR(MAX), 
    @query AS NVARCHAR(MAX) 

select @cols = STUFF((SELECT distinct ',' + QUOTENAME(statusname) 
        from assetstatus 
      FOR XML PATH(''), TYPE 
      ).value('.', 'NVARCHAR(MAX)') 
     ,1,1,'') 

set @query = 'SELECT typename, total,' + @cols + ' from 
      (
       select count(*) over(partition by t.typename) total, 
        s.statusname, 
        t.typename 
       from assets a 
       inner join assettypes t 
        on a.assettype = t.id 
       inner join assetstatus s 
        on a.assetstatus = s.id 
      ) x 
      pivot 
      (
       count(statusname) 
       for statusname in (' + @cols + ') 
      ) p ' 

execute(@query) 

参见SQL Fiddle with Demo

这还可以使用具有的情况下表达的聚集函数写为：

select typename, 
    total, 
    sum(case when statusname ='Deployed' then 1 else 0 end) Deployed, 
    sum(case when statusname ='Inventory' then 1 else 0 end) Inventory, 
    sum(case when statusname ='Shipped' then 1 else 0 end) Shipped 
from 
(
    select count(*) over(partition by t.typename) total, 
    s.statusname, 
    t.typename 
    from assets a 
    inner join assettypes t 
    on a.assettype = t.id 
    inner join assetstatus s 
    on a.assetstatus = s.id 
) d 
group by typename, total 

参见SQL Fiddle with Demo

---

MySQL的：此数据库做es没有支点函数，所以你将不得不使用聚合函数和CASE表达式。它也没有窗口函数，所以你将不得不稍微改变查询到以下几点：

select typename, 
    total, 
    sum(case when statusname ='Deployed' then 1 else 0 end) Deployed, 
    sum(case when statusname ='Inventory' then 1 else 0 end) Inventory, 
    sum(case when statusname ='Shipped' then 1 else 0 end) Shipped 
from 
(
    select t.typename, 
    (select count(*) 
    from assets a1 
    where a1.assettype = t.id 
    group by a1.assettype) total, 
    s.statusname 
    from assets a 
    inner join assettypes t 
    on a.assettype = t.id 
    inner join assetstatus s 
    on a.assetstatus = s.id 
) d 
group by typename, total; 

见SQL Fiddle with Demo

然后，如果你需要在MySQL中的动态解决方案，你将不得不使用准备语句生成SQL字符串来执行：

SET @sql = NULL; 
SELECT 
    GROUP_CONCAT(DISTINCT 
    CONCAT(
     'sum(CASE WHEN statusname = ''', 
     statusname, 
     ''' THEN 1 else 0 END) AS `', 
     statusname, '`' 
    ) 
) INTO @sql 
FROM assetstatus; 

SET @sql 
    = CONCAT('SELECT typename, 
       total, ', @sql, ' 
      from 
      (
       select t.typename, 
       (select count(*) 
       from assets a1 
       where a1.assettype = t.id 
       group by a1.assettype) total, 
       s.statusname 
       from assets a 
       inner join assettypes t 
       on a.assettype = t.id 
       inner join assetstatus s 
       on a.assetstatus = s.id 
      ) d 
      group by typename, total'); 

PREPARE stmt FROM @sql; 
EXECUTE stmt; 
DEALLOCATE PREPARE stmt; 

见SQL Fiddle with Demo。

结果是所有查询相同的两个数据库：

| TYPENAME | TOTAL | DEPLOYED | INVENTORY | SHIPPED | 
----------------------------------------------------- 
| Desktop |  2 |  1 |   1 |  0 | 
| Laptop |  1 |  0 |   0 |  1 | 
| Server |  1 |  1 |   0 |  0 |