Question

我有两个表，一个包含员工列表，另一个包含PO列表。

PO表如下：

PoID PO# Name             City     State
1    55  Jasons Company   Miami    FL
2    66  Mikes Company    NewYork  NY
3    77  Sallys Company   Dallas   TX

我的员工表看起来像：

EmployeeID  Name    Phone         Email    PoID
1           John    123-123-1233  j@j.com  1 
2           Alex    234-234-2344  a@a.com  1
3           Cindy   345-345-3455  c@c.com  2
4           Jessica 356-356-3566  j@j.com  3

现在我的SQL Select语句是这样的：

SELECT * FROM PO p
LEFT JOIN Employees e
ON p.PoID = e.PoID

它将返回我的PO表中的所有记录，也许只是与该记录关联的任何员工的最高条目。我还需要提取所有相关记录的名称/电子邮件。

我不确定如何实现这一目标，因为我不想要重复的PO条目，而只是一个以逗号分隔的列表，列出与该PO关联的任何可能的名称/电子邮件，以便在我的PO中添加两个额外的列

希望这是有道理的，如果有人遇到过这种情况，那么您的见解将受到赞赏。

Answer 1

您可以尝试此查询

SELECT p.*,
 STUFF((SELECT DISTINCT ',' + Name
          FROM @Employees e1
          WHERE e1.PoID = p.PoID
          FOR XML PATH (''))
          , 1, 1, '')  AS Names
,STUFF((SELECT DISTINCT ',' + Phone
          FROM @Employees e1
          WHERE e1.PoID = p.PoID
          FOR XML PATH (''))
          , 1, 1, '')  AS Phones
,STUFF((SELECT DISTINCT ',' + Email
          FROM @Employees e1
          WHERE e1.PoID = p.PoID
          FOR XML PATH (''))
          , 1, 1, '')  AS Emails
 FROM @PO p

结果

this explanation

下面您可以看到包含测试数据的整个脚本

declare @po table(PoID int, PO int, Name varchar(100),City varchar(100),State varchar(100))

insert into @po
select 1,55,'Jasons Company','Miami','FL' union
select 2,66,'Mikes Company','NewYork','NY' union
select 3,77,'Sallys Company','Dallas','TX'

declare @Employees table(EmployeeID int, Name varchar(100),Phone varchar(100),Email varchar(100), PoID int)

insert into @Employees
select 1,'John','123-123-1233','j@j.com',1  union
select 2,'Alex','234-234-2344','a@a.com',1 union
select 3,'Cindy','345-345-3455','c@c.com',2 union
select 4,'Jessica','356-356-3566','j@j.com',3

SELECT p.*,
 STUFF((SELECT DISTINCT ',' + Name
          FROM @Employees e1
          WHERE e1.PoID = p.PoID
          FOR XML PATH (''))
          , 1, 1, '')  AS Names
,STUFF((SELECT DISTINCT ',' + Phone
          FROM @Employees e1
          WHERE e1.PoID = p.PoID
          FOR XML PATH (''))
          , 1, 1, '')  AS Phones
,STUFF((SELECT DISTINCT ',' + Email
          FROM @Employees e1
          WHERE e1.PoID = p.PoID
          FOR XML PATH (''))
          , 1, 1, '')  AS Emails
 FROM @PO p

Answer 2

你可以使用它。

DECLARE @PO TABLE (PoID INT, PO INT,  Name VARCHAR(20), City VARCHAR(20), State vARCHAR(10))
INSERT INTO @PO
VALUES
(1, 55, 'Jasons Company','Miami','FL'),
(2, 66, 'Mikes Company','NewYork','NY'),
(3, 77, 'Sallys Company','Dallas','TX')

DECLARE @Employees  TABLE (EmployeeID INT, Name VARCHAR(20), Phone VARCHAR(20), Email VARCHAR(20), PoID INT)
INSERT INTO @Employees
VALUES
(1,'John','123-123-1233','j@j.com',1 ),
(2,'Alex','234-234-2344','a@a.com',1),
(3,'Cindy','345-345-3455','c@c.com',2),
(4,'Jessica','356-356-3566','j@j.com',3)



SELECT 
    *, 
    STUFF ( (SELECT 
            ', '  + Name + '/' + Email 
        FROM 
            @Employees e WHERE e.PoID = p.PoID FOR XML PATH('') ) ,1,1,'') [names/emails]
FROM 
    @PO p

结果

PoID        PO          Name                 City                 State      names/emails
----------- ----------- -------------------- -------------------- ---------- -------------------
1           55          Jasons Company       Miami                FL          John/j@j.com, Alex/a@a.com
2           66          Mikes Company        NewYork              NY          Cindy/c@c.com
3           77          Sallys Company       Dallas               TX          Jessica/j@j.com

TSQL从两个非奇偶校验表中提取记录

2 个答案: