Question

我需要帮助创建以下结果。我想到了一个sql pivot，但我不知道如何使用它。看了几个例子，无法想出解决方案。关于如何实现这一目标的任何其他想法也是受欢迎的。必须动态生成状态列。

有三个表，资产，资产类型，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

这种类型的转换称为枢轴。您没有指定您正在使用的数据库，因此我将为SQL Server和MySQL提供答案。

SQL Server：如果您使用的是SQL Server 2005+，则可以实现PIVOT功能。

如果您要将已知数量的值转换为列，则可以对查询进行硬编码。

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。

但是如果你有一个未知数量的status值，那么你需要使用动态sql在运行时生成列列表。

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)

请参阅SQL Fiddle with Demo

这也可以使用带有case表达式的聚合函数来编写：

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

请参阅SQL Fiddle with Demo

MySQL：此数据库没有 pivot 功能，因此您必须使用聚合函数和CASE表达式。它也没有窗口函数，因此您必须稍微更改查询到以下内容：

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;

请参阅SQL Fiddle with Demo

然后，如果您需要MySQL中的动态解决方案，则必须使用预准备语句来生成要执行的sql字符串：

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;

请参阅SQL Fiddle with Demo。

两个数据库中的所有查询结果相同：

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

Answer 2

使用不兼容数据透视表的DBMS（绝对数据库），使用此SQL交叉表等效语句更加成功：

SELECT
  sub.TypeName
, SUM(sub.[Count]) AS "Total"
, SUM(CASE WHEN AssetStatus='1' THEN sub.[Count] ELSE 0 END) AS "Deployed"
, SUM(CASE WHEN AssetStatus='2' THEN sub.[Count] ELSE 0 END) AS "Inventory"
, SUM(CASE WHEN AssetStatus='3' THEN sub.[Count] ELSE 0 END) AS "Shipped"
FROM
 (
SELECT
  t.TypeName
, AssetStatus
, COUNT(AssetID) AS "Count"
FROM
  Assets
  JOIN AssetTypes t ON t.ID = AssetType
  JOIN AssetStatus s ON s.ID = AssetStatus
GROUP BY t.TypeName, AssetStatus, s.StatusName
 ) sub
GROUP BY sub.TypeName
;

当我意识到该代码（上述）不适用于MySQL时，我修改了以下代码，使其在MySQL中的性能与在当前的Absolute Database中一样好。原因是特定的NULL处理避免了dBase，悖论以及绝对接受主流数据库中不接受的COUNT（NULL）= 0的绝对数据库的陷阱。因此，相信这将在大多数数据库中很好地执行（处理CASE ..），这是我改编的代码：

SELECT
  sub.TypeName
, SUM(sub.AssetCase) AS "Total"
, SUM(CASE WHEN sub.StatusName = 'Deployed' THEN sub.AssetCase ELSE 0 END) AS "Deployed"
, SUM(CASE WHEN sub.StatusName = 'Inventory' THEN sub.AssetCase ELSE 0 END) AS "Inventory"
, SUM(CASE WHEN sub.StatusName = 'Shipped' THEN sub.AssetCase ELSE 0 END) AS "Shipped"
FROM
  (
   SELECT
     c.TypeName
   , c.StatusName
   , CASE WHEN a.AssetID IS NULL THEN 0 ELSE 1 END AS "AssetCase"
   FROM
     (
      SELECT
        t.ID AS tID
      , t.TypeName
      , s.ID AS sID
      , s.StatusName
      FROM
        AssetTypes t, AssetStatus s
     ) c
   LEFT JOIN Assets a
     ON a.AssetType = c.tID AND a.AssetStatus = c.sID
   ) sub
GROUP BY
  sub.TypeName
;

最佳问候 Niels Knabe

我需要知道如何创建交叉表查询

2 个答案: