生成唯一标识符作为基于集合的查询

Question

我正在从完全基于光标的基础集合移动一堆代码并生成这一直在做我的头。我们创建一个6字符的短代码（唯一）插入到每个公司数据库和我（想）在游标之外实现这一点。的我在迄今为止其中

例子：

CREATE TABLE #customers (name VARCHAR(50), shortname VARCHAR(10)) 
INSERT INTO #customers VALUES 
('Michael Smith', 'Michae') 
,('Michael Douglas', 'Mich_1') 
,('Michael Yang', 'Mich_2') 

CREATE TABLE #newcustomers (name VARCHAR(50), shortname VARCHAR(10) NULL) 
INSERT INTO #newcustomers (name) VALUES 
('Michael Black') 
,('Michael White') 

SELECT * FROM #customers 
SELECT * FROM #newcustomers 

DECLARE @shortname VARCHAR(10) 
DECLARE @iteration INT = 0 

WHILE EXISTS(SELECT shortname FROM #customers WHERE shortname = @shortname) 
BEGIN 
    SELECT @shortname = LEFT(name, 6) FROM #newcustomers 

    UPDATE #newcustomers SET shortname = @shortname 

    SET @shortname = LEFT(@shortname, 4) + '_' + @iteration 

    SET @iteration = @iteration + 1 
END 

希望的例子是在查明我试图去，有什么建议或例子将是非常有益足够了。我的例子不起作用。

Answer 1

试试这个

你的表作为实物模型

CREATE TABLE #customers (ID INT IDENTITY, name VARCHAR(50), shortname VARCHAR(10)) 
INSERT INTO #customers VALUES 
('Michael Smith', 'Michae') 
,('Michael Douglas', 'Mich_1') 
,('Michael Yang', 'Mich_3') 
,('Testman', 'Testma') 
,('Testman1', 'Test_1'); 

CREATE TABLE #newcustomers (ID INT IDENTITY,name VARCHAR(50), shortname VARCHAR(10) NULL) 
INSERT INTO #newcustomers (name) VALUES 
('Michael Black') 
,('Michael White') 
,('Testman2') 
,('Someone new'); 

--This CTE将结合现有的所有名称

WITH AllNames AS 
(
    SELECT '1_old' AS datasource,ID,name,shortname FROM #customers 
    UNION ALL SELECT '2_new',ID,name,shortname FROM #newcustomers 
) 

--This CTE将使用组合列表并计算出正确的“索引”

,ShortNames AS 
(
    SELECT c.* 
      ,A.First6 
      ,ROW_NUMBER() OVER(PARTITION BY A.First6 ORDER BY datasource,ID) AS NrTotal 
      ,ROW_NUMBER() OVER(PARTITION BY datasource,A.First6 ORDER BY datasource,ID) AS Nr 
      ,CASE WHEN ISNUMERIC(SUBSTRING(shortname+'  ',6,10))=1 
       THEN CAST(SUBSTRING(shortname+'  ',6,10) AS INT) ELSE 0 END AS ExistIndex 
    FROM AllNames AS c 
    CROSS APPLY(SELECT LEFT(name + '  ',6)) AS A(First6) 
) 

- 所有新的与NrTotal = 1获得6个字母的是，所有其他的景气指数

SELECT * 
     ,CASE WHEN datasource='1_old' THEN shortname ELSE 
     CASE WHEN datasource='2_new' AND NrTotal=1 THEN First6 
      ELSE LEFT(First6,4) + '_' + CAST(Nr + (SELECT ISNULL(MAX(x.ExistIndex),1) 
                 FROM ShortNames AS x 
                 WHERE x.First6=ShortNames.First6) AS VARCHAR(5)) 
     END 
     END 
FROM ShortNames 

GO 
DROP TABLE #customers; 
DROP TABLE #newcustomers; 

结果

+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| datasource | ID | name   | shortname | First6 | NrTotal | Nr | ExistIndex | (Kein Spaltenname) | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 1_old  | 1 | Michael Smith | Michae | Michae | 1  | 1 | 0   | Michae    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 1_old  | 2 | Michael Douglas | Mich_1 | Michae | 2  | 2 | 1   | Mich_1    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 1_old  | 3 | Michael Yang | Mich_3 | Michae | 3  | 3 | 3   | Mich_3    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 1_old  | 4 | Testman   | Testma | Testma | 1  | 1 | 0   | Testma    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 1_old  | 5 | Testman1  | Test_1 | Testma | 2  | 2 | 1   | Test_1    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 2_new  | 1 | Michael Black | NULL  | Michae | 4  | 1 | 0   | Mich_4    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 2_new  | 2 | Michael White | NULL  | Michae | 5  | 2 | 0   | Mich_5    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 2_new  | 4 | Someone new  | NULL  | Someon | 1  | 1 | 0   | Someon    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 
| 2_new  | 3 | Testman2  | NULL  | Testma | 3  | 1 | 0   | Test_2    | 
+------------+----+-----------------+-----------+--------+---------+----+------------+--------------------+ 

Answer 2

一种选择是使用computed column。

沿着这些线路的表设计将工作：

- Sample table. 
DECLARE @Sample TABLE 
    (
     Id   INT IDENTITY(1, 1), 
     FullName VARCHAR(255), 
     ShortName AS LEFT(FullName, 4) + '_' + CAST(Id AS VARCHAR(255)) 
    ) 
; 

-- Add set containing two companies. 
INSERT INTO @Sample 
    (
     FullName 
    ) 
VALUES 
    ('ABC LTD'), 
    ('XYZ PLC') 
; 

返回

Id FullName ShortName 
1 ABC LTD  ABC _1 
2 XYZ PLC  XYZ _1 

的Id和ShortName列将通过SQL Server管理。您只需要添加FullName。

编辑

使用表变量，以使其更容易发挥沿

返工例子。