SQL转置和添加列

Posted 2021-04-16

我有一个具有以下结构（标题）的表：

ProjID,Cost2001,Cost2002,Cost2003

例如，前两行看起来像这样：

projectA,10,32,30
projectB,42,22,122

我想将此表转换为以下结构（标题）：

ProjID,CostYear,Value

因此回到示例数据，发布转换，它看起来像这样：

ProjectA,Cost2001,10
ProjectA,Cost2002,32
ProjectA,Cost2003,30
ProjectB,Cost2001,42
ProjectB,Cost2002,22
ProjectB,Cost2003,122

我该怎么做呢？我正在使用支持Standard SQL的Google BigQuery。我只需要这样做一次来修复表，所以我不介意将数据导入另一个RDBMS以便能够使用数据透视功能。

答案

恕我直言，最简单的方法是使用UNION ALL。

CREATE TABLE projects(ProjID VARCHAR(20), Cost2001 int, Cost2002 int, Cost2003 int);
INSERT INTO projects VALUES
('projectA', 10, 32, 30),
('projectB', 42, 22, 122);

CREATE TABLE new_projects (ProjID VARCHAR(20), ProjYear INT, Cost int);
GO

2 rows affected

INSERT INTO new_projects
SELECT ProjID, 2001, Cost2001
FROM   projects
UNION ALL
SELECT ProjID, 2002, Cost2002
FROM   projects
UNION ALL
SELECT ProjID, 2003, Cost2003
FROM   projects;

SELECT * FROM new_projects;
GO

ProjID   | ProjYear | Cost
:------- | -------: | ---:
projectA |     2001 |   10
projectB |     2001 |   42
projectA |     2002 |   32
projectB |     2002 |   22
projectA |     2003 |   30
projectB |     2003 |  122

dbfiddle here

另一答案

以下是真正的BigQuery风格：o）

这两个版本都适用于BigQuery Standard SQL

##standardSQL
SELECT 
  projID,  
  ([2001, 2002, 2003])[SAFE_OFFSET(pos)] year, 
  cost
FROM `project.dataset.table`,
UNNEST([Cost2001,Cost2002,Cost2003]) cost WITH OFFSET pos

您可以使用您问题中的虚拟数据测试/播放上面的智慧，如下所示

##standardSQL
WITH `project.dataset.table` AS (
  SELECT 'projectA' projID, 10 Cost2001, 32 Cost2002, 30 Cost2003 UNION ALL
  SELECT 'projectB', 42, 22, 122 
)
SELECT 
  projID,  
  ([2001, 2002, 2003])[SAFE_OFFSET(pos)] year, 
  cost
FROM `project.dataset.table`,
UNNEST([Cost2001,Cost2002,Cost2003]) cost WITH OFFSET pos   

结果为

Row projID      year    cost     
1   projectA    2001    10   
2   projectA    2002    32   
3   projectA    2003    30   
4   projectB    2001    42   
5   projectB    2002    22   
6   projectB    2003    122    

正如您在上面的查询中所看到的，您必须在下面的行中预先设置各年的值

([2001, 2002, 2003])[SAFE_OFFSET(pos)] year   

如果出于某种原因你想要更通用并且能够从原始列的名称中获得这些值 - 你可以使用以下通用方法

##standardSQL
WITH `project.dataset.table` AS (
  SELECT 'projectA' projID, 10 Cost2001, 32 Cost2002, 30 Cost2003 UNION ALL
  SELECT 'projectB', 42, 22, 122 
)
SELECT 
  projID,  
  SPLIT(x,':')[SAFE_OFFSET(0)] year,
  SPLIT(x,':')[SAFE_OFFSET(1)] cost 
FROM `project.dataset.table` t,
UNNEST(SPLIT(REGEXP_REPLACE(TO_JSON_STRING(t), r'[{}"]', ''))) x
WHERE SPLIT(x,':')[OFFSET(0)] != 'projID'

显然，结果相同

Row projID      year        cost     
1   projectA    Cost2001    10   
2   projectA    Cost2002    32   
3   projectA    Cost2003    30   
4   projectB    Cost2001    42   
5   projectB    Cost2002    22   
6   projectB    Cost2003    122  

另一答案

正如大家所提到的，你需要UNPIVOT，这看起来像：

DECLARE @projects TABLE (projid nvarchar(max), cost2001 int, cost2002 int, cost2003 int);

INSERT @projects VALUES ('projectA', 10, 32, 30)
                    , ('projectB', 42, 22, 122);


SELECT PROJID, PROJECT_ATTRIBUTE, PROJECT_COST
FROM @projects
UNPIVOT (PROJECT_COST FOR PROJECT_ATTRIBUTE in (cost2001, cost2002, cost2003) ) AS UNPVT

出于性能原因，我不打算使用UNION ALL版本。基本上你会扫描表3次或者你有多少“CostYear”列，而且你必须为它添加一个全新的查询。

而不是用UNPIVOT扫描表一次。

另一答案

取消您的数据

DECLARE @ProjectTbl TABLE (ProjectID VARCHAR(225),Cost2001 INT, Cost2002 INT,Cost2003  INT)
INSERT INTO @ProjectTbl VALUES
('projectA',10,32,30),
('projectB',42,22,122);

;WITH Unpivots 
AS
(SELECT 
*
FROM @ProjectTbl
UNPIVOT 
(
  Value FOR CostYear IN (Cost2001, Cost2002, Cost2003)
) AS up 
)
SELECT
ProjectID,
CostYear,
Value
FROM Unpivots

产量

ProjectID   CostYear    Value
projectA    Cost2001    10
projectA    Cost2002    32
projectA    Cost2003    30
projectB    Cost2001    42
projectB    Cost2002    22
projectB    Cost2003    122