Question

我有两个表：

TableA，其中包含所有产品的说明：

codeProduct   description

    1              ok
    2              yes

TableB仅包含具有以下代码的产品之间的层次结构：

level_1 level_2 level_3 level_4
1       2       23      75
1       2       53      85

我如何获得包含每个级别描述的最终表

level_1 description_1 level_2 description_2 level_3 description_3 level_4 description_4

Answer 1

每次需要tableB中的列的值时，都需要多次使用tableA

select b.level_1
    , a1.description description_1
    , b.level_2
    , a2.description description_2
    , b.level_3
    , a3.description description_3
    , b.level_4 
    , a4.description description_4 
from  TableB b 
left join  TableA a1 on a1.codeProduct = b.level_1 
left join  TableA a2 on a2.codeProduct = b.level_2 
left join  TableA a3 on a3.codeProduct = b.level_3 
left join  TableA a4 on a4.codeProduct = b.level_4

如果两个表之间的值都不匹配，则使用左连接；如果您拥有所有的值，则使用INNER JOIN

select b.level_1
    , a1.description description_1
    , b.level_2
    , a2.description description_2
    , b.level_3
    , a3.description description_3
    , b.level_4 
    , a4.description description_4 
from  TableB b 
INNER join  TableA a1 on a1.codeProduct = b.level_1 
INNER join  TableA a2 on a2.codeProduct = b.level_2 
INNER join  TableA a3 on a3.codeProduct = b.level_3 
INNER join  TableA a4 on a4.codeProduct = b.level_4

Answer 2

尝试一下：

CREATE TABLE #TABLEA
(
  CODEPRODUCT INT NOT NULL
, DESCRIPTION VARCHAR (100) NOT NULL
);

INSERT INTO #TABLEA VALUES (1, 'ok') ;
INSERT INTO #TABLEA VALUES (2, 'yes');

CREATE TABLE #TABLEB
(
  LEVEL_1 INT NOT NULL
, LEVEL_2 INT NOT NULL
, LEVEL_3 INT NOT NULL
, LEVEL_4 INT NOT NULL
);

INSERT INTO #TABLEB VALUES(1, 2, 23, 75) , (1, 2, 53, 85);

SELECT #TABLEB.LEVEL_1
     , TA_L1.DESCRIPTION AS DESCRIPTION1
     , #TABLEB.LEVEL_2
     , TA_L2.DESCRIPTION AS DESCRIPTION2
     , #TABLEB.LEVEL_3
     , TA_L3.DESCRIPTION AS DESCRIPTION3
     , #TABLEB.LEVEL_4
     , TA_L4.DESCRIPTION AS DESCRIPTION4
FROM #TABLEB
LEFT JOIN #TABLEA TA_L1
ON #TABLEB.LEVEL_1 = TA_L1.CODEPRODUCT
LEFT JOIN #TABLEA TA_L2
ON #TABLEB.LEVEL_2 = TA_L2.CODEPRODUCT
LEFT JOIN #TABLEA TA_L3
ON #TABLEB.LEVEL_3 = TA_L3.CODEPRODUCT
LEFT JOIN #TABLEA TA_L4
ON #TABLEB.LEVEL_4 = TA_L4.CODEPRODUCT;

结果：

LEVEL_1, DESCRIPTION1, LEVEL_2, DESCRIPTION2, LEVEL_3, DESCRIPTION3, LEVEL_4, DESCRIPTION4
    1   ok  2   yes 23  NULL    75  NULL
    1   ok  2   yes 53  NULL    85  NULL

两个表之间的链接

2 个答案: