Question

我正在寻找最干净的SQL查询来实现以下目标。性能不是那么重要，因为我的数据集很小。

Sample table:
Letter field holding: A, B, C, D, E,
Location field holding: UAE, CANADA, BOSTON, BAHRAIN, FRANCE

我正在寻找一个结果，列出其中包含字母组合的每个字母/位置，因此结果如下：

A-UAE
A-CANADA
A-BAHRAIN
A-FRANCE
B-BOSTON
B-BAHRAIN
C-CANADA
C-FRANCE
D-CANADA
E-UAE
E-FRANCE

Answer 1

这是另一种解决方案：

DECLARE @Letter TABLE (
  letter CHAR(1) PRIMARY KEY
  );

DECLARE @Country TABLE (
  name VARCHAR(100) PRIMARY KEY
  );

INSERT INTO @Letter (letter)
VALUES ('A'), ('B'), ('C'), ('D'), ('E');

INSERT INTO @Country (name)
VALUES ('UAE'), ('CANADA'), ('BOSTON'), ('BAHRAIN'), ('FRANCE');

SELECT CONCAT(L.letter, ' - ', C.name)
FROM @Letter AS L
INNER JOIN @Country AS C
  ON C.name LIKE '%' + L.letter + '%'
ORDER BY L.letter, C.name;

结果：

A - BAHRAIN 
A - CANADA  
A - FRANCE  
A - UAE     
B - BAHRAIN 
B - BOSTON  
C - CANADA  
C - FRANCE  
D - CANADA  
E - FRANCE  
E - UAE

希望你输出你期望的东西您可以在堆栈交换数据上运行此查询：https://data.stackexchange.com/stackoverflow/query/622821

或者，如果性能出现问题，您可以创建一个单独的表格，用于存储每个国家/地区名称及其名称。唯一的字母，所以你可以做一个简单的连接，而不是LIKE来比较事情：

DECLARE @CountrySplit TABLE (
  letter CHAR(1)
  , name VARCHAR(100)
  , PRIMARY KEY (letter, name)
  );

INSERT INTO @CountrySplit (letter, name)
SELECT DISTINCT SUBSTRING(C.name, v.number + 1, 1), C.name
FROM @Country AS C
INNER JOIN master..spt_values AS V
    ON V.number < LEN(C.name)
WHERE V.type = 'P';

SELECT CONCAT(L.letter, ' - ', CS.name) AS Result
FROM @CountrySplit AS CS
INNER JOIN @Letter AS L
    ON L.letter = CS.letter;

这是对堆栈交换数据的查询： https://data.stackexchange.com/stackoverflow/query/622841

对此字符串拆分答案的肯定：T-SQL Split Word into characters

Answer 2

借助Parse / Split UDF和Cross Apply。

我添加了一个ID来证明可以为整个表运行

示例

Declare @YourTable table (ID int,Letter varchar(50),Location varchar(50)) Insert Into @YourTable values (1,'A, B, C, D, E,','UAE, CANADA, BOSTON, BAHRAIN, FRANCE') Select A.ID ,B.* From @YourTable A Cross Apply ( Select NewValue = B1.RetVal+'-'+B2.RetVal From [dbo].[udf-Str-Parse](A.Letter,',') B1 Join [dbo].[udf-Str-Parse](A.Location,',') B2 on charindex(B1.RetVal,B2.RetVal)>0 ) B

<强>返回

ID NewValue 1 A-UAE 1 A-CANADA 1 A-BAHRAIN 1 A-FRANCE 1 B-BOSTON 1 B-BAHRAIN 1 C-CANADA 1 C-FRANCE 1 D-CANADA 1 E-UAE 1 E-FRANCE

UDF（如果需要）

CREATE FUNCTION [dbo].[udf-Str-Parse] (@String varchar(max),@Delimiter varchar(10)) Returns Table As Return ( Select RetSeq = Row_Number() over (Order By (Select null)) ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)'))) From (Select x = Cast('<x>' + replace((Select replace(@String,@Delimiter,'§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A Cross Apply x.nodes('x') AS B(i) ); --Thanks Shnugo for making this XML safe --Select * from [dbo].[udf-Str-Parse]('Dog,Cat,House,Car',',') --Select * from [dbo].[udf-Str-Parse]('John Cappelletti was here',' ') --Select * from [dbo].[udf-Str-Parse]('this,is,<test>,for,< & >',',')

编辑 - 没有UDF的选项

Declare @YourTable table (ID int,Letter varchar(50),Location varchar(50)) Insert Into @YourTable values (1,'A, B, C, D, E,','UAE, CANADA, BOSTON, BAHRAIN, FRANCE') Select A.ID ,B.* From @YourTable A Cross Apply ( Select NewValue = B1.RetVal+'-'+B2.RetVal From ( Select RetSeq = Row_Number() over (Order By (Select null)) ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)'))) From (Select x = Cast('<x>' + replace((Select replace(A.Letter,',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A Cross Apply x.nodes('x') AS B(i) ) B1 Join ( Select RetSeq = Row_Number() over (Order By (Select null)) ,RetVal = LTrim(RTrim(B.i.value('(./text())[1]', 'varchar(max)'))) From (Select x = Cast('<x>' + replace((Select replace(A.Location,',','§§Split§§') as [*] For XML Path('')),'§§Split§§','</x><x>')+'</x>' as xml).query('.')) as A Cross Apply x.nodes('x') AS B(i) ) B2 on charindex(B1.RetVal,B2.RetVal)>0 ) B

编写递归SQL视图的最佳方法

2 个答案: