计算匹配值并以矩阵形式显示答案

【问题标题】：Count the match values and display in matrix form计算匹配值并以矩阵形式显示
【发布时间】：2019-01-10 11:36:39
【问题描述】：

我有以下表格，其中包含一些示例数据：

表：TblTestDB

CREATE TABLE TblTestDB (id int,name varchar(100));

INSERT INTO TblTestDB VALUES(1,'Sam'),(2,'Jack'),
                            (3,'Rock'),(4,'Don'),(5,'Tam');

表：TblDB1

CREATE TABLE TblDB1 (id int,name varchar(100));

INSERT INTO TblDB1 VALUES(1,'Sam'),(2,'Jack'),
                         (3,'Rock');

表：TblDB2

CREATE TABLE TblDB2 (id int,name varchar(100));

INSERT INTO TblDB2 VALUES(1,'Jazz'),(2,'Dsouze'),
                         (3,'Rock'),(4,'Jack');

表：TblDB3

CREATE TABLE TblDB3 (id int,name varchar(100));

INSERT INTO TblDB3 VALUES(1,'Sam'),(2,'Jazz'),
                         (3,'Rock');

我想以如下形式显示结果：

TableName   Name    ID
------------------------------
TblDB1      3       3
TblDB2      2       4
TblDB3      2       3

关于结果集的说明：我想显示在表TblTestDB 和其他表（TblDB1,TblDB2,TblDB3）之间匹配的列值的计数。

尝试使用以下查询：

查询：

SELECT DB,MAX(Name) AS Name, MAX(ID) AS ID
FROM
(
    SELECT 'TblDB1' AS DB,COUNT(a.Name) AS Name,0 AS ID 
    FROM TblTestDB a
    INNER JOIN TblDB1 b ON a.Name = b.Name
    UNION
    SELECT 'TblDB2' AS DB,COUNT(a.Name) AS Name,0 AS ID 
    FROM TblTestDB a
    INNER JOIN TblDB2 b ON a.Name = b.Name
    UNION
    SELECT 'TblDB3' AS DB,COUNT(a.Name) AS Name,0 AS ID 
    FROM TblTestDB a
    INNER JOIN TblDB3 b ON a.Name = b.Name
    UNION
    SELECT 'TblDB1' AS DB,0 AS Name,COUNT(a.ID) AS ID 
    FROM TblTestDB a
    INNER JOIN TblDB1 b  ON a.ID = b.ID
    UNION
    SELECT 'TblDB2' AS DB,0 AS Name,COUNT(a.ID) AS ID 
    FROM TblTestDB a
    INNER JOIN TblDB2 b  ON a.ID = b.ID
    UNION
    SELECT 'TblDB3' AS DB,0 AS Name,COUNT(a.ID) AS ID 
    FROM TblTestDB a
    INNER JOIN TblDB3 b ON a.ID = b.ID
) a
GROUP BY DB

问题：我可能会得到超过 10 列来显示，如矩阵/枢轴，上面的查询根据列列表增长。任何更好的方法来使它简短而简单。

【问题讨论】：

标签： sql-server tsql sql-server-2008-r2

【解决方案1】：

SELECT DISTINCT 'TblDB1' AS DB
      ,COUNT(CASE WHEN a.Name = b.Name THEN 1 ELSE NULL END) AS Name
      ,COUNT(CASE WHEN a.ID = b.ID THEN 1 ELSE NULL END) AS ID 
FROM TblTestDB a
CROSS JOIN TblDB1 b
UNION ALL
SELECT DISTINCT 'TblDB2' AS DB
      ,COUNT(CASE WHEN a.Name = b.Name THEN 1 ELSE NULL END) AS Name
      ,COUNT(CASE WHEN a.ID = b.ID THEN 1 ELSE NULL END) AS ID 
FROM TblTestDB a
CROSS JOIN TblDB2 b
UNION ALL
SELECT DISTINCT 'TblDB3' AS DB
      ,COUNT(CASE WHEN a.Name = b.Name THEN 1 ELSE NULL END) AS Name
      ,COUNT(CASE WHEN a.ID = b.ID THEN 1 ELSE NULL END) AS ID 
FROM TblTestDB a
CROSS JOIN TblDB3 b

【讨论】：

【解决方案2】：

这么简单吗？

SELECT 
    tableName, sum(s.[Name]) as ncnt, max(s.id) as ID
FROM TblTestDB t
cross apply (
    SELECT tableName, count(name) as [name], max(id) as ID
    FROM (
        SELECT 'TblDB1' as tid, * FROM TblDB1
        union
        SELECT 'TblDB2' as tid, * FROM TblDB2
        union
        SELECT 'TblDB3' as tid, * FROM TblDB3
    ) i
    where i.name = t.name
    group by tableName
) s
group by tableName

不知道你的意思是什么，“我可能会得到超过 10 列来显示像矩阵/枢轴”。例如，如果您要添加一个姓氏列，应该独立匹配，还是与名称列结合？（前者可能有点复杂……）

【讨论】：