Question

当我执行以下代码时，我得到的结果如下：

ID  column1 column2 

34  NULL    NULL
34  Org13   Org13
36  NULL    NULL
36  NULL    Org2
36  Org4    NULL
41  NULL    NULL
41  NULL    Org5
41  Org3    NULL

我希望我的结果看起来像：

ID  column1  column2

34  Org13   Org13
36  Org4    Org2
41  Org3    Org5

我有两个表：Table1和Table2。 Table2是一个包含以下字段的查找表：id，name

Table1具有以下字段（id，column1，column2）。 column1和column2都具有与查找表的外键关系：

FK_1: Table1.column1-Table2.id
FK_2: Table1.column2-Table2.id

因为我想拉出column1和column2的值，并且因为这两个值都是在同一个字段（Table2.name）上查找，所以我怀疑我需要做内部选择。

我的代码如下。我怎样才能改变它，以便产生所需的结果，而不是我得到的结果呢？提前谢谢！

DECLARE @value INT
SET @value = 14

SELECT DISTINCT 
    Table1.[id]         AS ID
    , ( SELECT DISTINCT
            Table2.[name] 
        WHERE 
            Table1.column1 =
            Table2.id ) AS column1
    , ( SELECT DISTINCT
            Table2.[name] 
        WHERE 
            Table1.column2 =
            Table2.id ) AS column2
FROM 
    Table1
    ,Table2
WHERE   
    Table1.[id] = @value

Answer 1

    /*
    create table table1(id int, col1 int, col2 int);
    create table table2(id int, name varchar(10) );

    insert into table2 values(1, 'org 1');
    insert into table2 values(2, 'org 2');
    insert into table2 values(3, 'org 3');
    insert into table2 values(4, 'org 4');

    insert into table1 values(1, 1, 2);
    insert into table1 values(2, 2, 2);
    insert into table1 values(3, 2, 3);
    insert into table1 values(4, 4, 1);
    */

    select
        a.id,
        b.name as column1,
        c.name as column2
    from
         table1 a
    join table2 b on b.id = a.col1
    join table2 c on c.id = a.col2;


 id     column1     column2    
 -----  ----------  ---------- 
 1      org 1       org 2      
 2      org 2       org 2      
 3      org 2       org 3      
 4      org 4       org 1      

 4 record(s) selected [Fetch MetaData: 3/ms] [Fetch Data: 0/ms] 

 [Executed: 7/7/09 4:07:25 PM EDT ] [Execution: 1/ms]

Answer 2

DECLARE @value INT
SET @value = 14

SELECT
    t1.[id]                 AS ID
    MAX(t2a.name),
    MAX(t2b.name)
FROM 
    Table1 t1
    LEFT JOIN
    Table2 t2a ON t1.column1 = t2a.id
    LEFT JOIN
    Table2 t2b ON t1.column2 = t2b.id
WHERE   
    t1.[id] = @value
GROUP BY    
    t1.[id]

Answer 3

gbn，我想你打算写

DECLARE @value INT
SET @value = 1

SELECT --??? DISTINCT 
    t1.[id] AS ID, --- missed comma
    table2a.name,
    table2b.name
FROM 
   Table1 t1
     JOIN Table2 table2a ON t1.column1 = table2a.id
     JOIN Table2 table2b ON t1.column2 = table2b.id -- you have t1.column1 oops
WHERE   
    t1.[id] = @value

如何在此代码中正确构造SQL子查询？

3 个答案: