Question

我的SQL表就像下面的

City_Code     Post_Code    Post_Code_Description
100           A1           ABC
100           C8           XYZ
100           Z3           MNO
200           D4           LMN
300           E3           IJK
300           B9           RST

这是city_code和Post_Code之间的映射。一个City_Code有很多邮政编码。现在我想运行一个Query来获得类似下面的内容

City_Code     Post_Code    Post_Code_Description
100           A1,C8,Z3     ABC,XYZ,MNO
200           D4           LMN
300           E3,B9        IJK,RST

你能帮我解决一下这个SQL表如下吗

Answer 1

试试这个：

SELECT City_Code, 
      Post_Code = 
        STUFF((SELECT ', ' + Post_Code
           FROM your_table b 
           WHERE b.City_Code = a.City_Code 
          FOR XML PATH('')), 1, 2, ''),
      Post_Code_Description=
        STUFF((SELECT ', ' + Post_Code_Description
           FROM your_table b 
           WHERE b.City_Code = a.City_Code 
          FOR XML PATH('')), 1, 2, '')
FROM your_table a
GROUP BY City_Code

Answer 2

如果您使用的是MySQL，可以使用GROUP_CONCAT()

select City_Code, 
     GROUP_CONCAT(Post_Code) Post_Code, 
     GROUP_CONCAT(Post_Code_Description) post_code_description
from yourtable
group by City_Code

对于SQL Server，您可以使用STUFF()和FOR XML PATH()

select city_code,
    Stuff((SELECT ', ' + post_code 
            FROM yourtable t2
            where t1.city_code = t2.city_code
            FOR XML path('')),1,1,'') Post_Code,
    Stuff((SELECT ', ' + post_code_description
            FROM yourtable t2
            where t1.city_code = t2.city_code
            FOR XML path('')),1,1,'') post_code_description
from yourtable t1
group by city_code

Answer 3

试试这个：

select city_code,substring((select ',' + post_code  
from city b where a.city_code=b.city_code
for xml path('')
),2,100)as post_code,
substring((select ',' + post_code_description 
from city c where a.city_code=c.city_code
for xml path('')
),2,100)as post_code_description
from city a
group by a.city_code

Answer 4

对此使用递归查询：

--Prepare Dummy Data
;WITH CITIES 
     AS (SELECT 100   AS City_Code, 
                'A1'  AS Post_code, 
                'ABC' AS Post_Code_Description 
         UNION 
         SELECT 100   AS City_Code, 
                'C8'  AS Post_code, 
                'XYZ' AS Post_Code_Description 
         UNION 
         SELECT 100   AS City_Code, 
                'Z3'  AS Post_code, 
                'MNO' AS Post_Code_Description 
         UNION 
         SELECT 200   AS City_Code, 
                'D4'  AS Post_code, 
                'LMN' AS Post_Code_Description 
         UNION 
         SELECT 300   AS City_Code, 
                'E3'  AS Post_code, 
                'IJK' AS Post_Code_Description 
         UNION 
         SELECT 300   AS City_Code, 
                'B9'  AS Post_code, 
                'RST' AS Post_Code_Description), 
--Add Row numbers to each row
     PREPARE 
     AS (SELECT *, 
                ROW_NUMBER () 
                  OVER ( 
                    PARTITION BY CITY_CODE 
                    ORDER BY CITY_CODE) RN 
         FROM   CITIES),
--Start Recursive CTE 
     RECURSIVE 
     AS (
--Anchor Query
         SELECT CITY_CODE, 
                CAST(POST_CODE AS VARCHAR(MAX))             Post_code, 
                CAST(POST_CODE_DESCRIPTION AS VARCHAR(MAX)) 
                Post_Code_Description, 
                1                                           AS LEVEL, 
                RN 
         FROM   PREPARE 
         WHERE  RN = 1 
         UNION ALL 
--Recursive Query
         SELECT T1.CITY_CODE, 
                T1.POST_CODE + ',' + T2.POST_CODE, 
                T1.POST_CODE_DESCRIPTION + ',' 
                + T2.POST_CODE_DESCRIPTION, 
                T2.LEVEL + 1, 
                T1.RN 
         FROM   PREPARE AS T1 
                INNER JOIN RECURSIVE AS T2 
                        ON T1.RN = T2.RN + 1 
                           AND T1.CITY_CODE = T2.CITY_CODE) 
--Final Results
SELECT T1.CITY_CODE, 
       T1.POST_CODE, 
       T1.POST_CODE_DESCRIPTION 
FROM   RECURSIVE T1 
       INNER JOIN (SELECT CITY_CODE, 
                          COUNT(*) cnt 
                   FROM   CITIES 
                   GROUP  BY CITY_CODE)T2 
               ON T1.CITY_CODE = T2.CITY_CODE 
WHERE  T1.LEVEL = T2.CNT

逗号使用SQL Query分隔值

4 个答案: