SQL基础知识--多行查询结果拼接

 1 select PERSON_NAME from S_PROJECT_MEMBER where PROJECT_ID = ‘2c9081473e2f8bda013e3028e4700049‘;  

先直接上代码吧！-------------------------有问题的，麻烦指出--------------------------------------------------------

select  
    CAST(stuff((
        select distinct ‘,‘+PERSON_NAME 
        from S_PROJECT_MEMBER  
        where PROJECT_ID = ‘2c9081473e2f8bda013e3028e4700049‘ for xml path(‘‘)),1,1,‘‘) AS varchar(1000)
    ) 
        AS PERSON_NAMES 
from S_PROJECT_MEMBER 
where PROJECT_ID = ‘2c9081473e2f8bda013e3028e4700049‘ GROUP BY PROJECT_ID; 

这里用的知识点有：

for xml path() 函数：

--这里用来做字符串拼接

CAST()函数:

--CAST函数用于将某种数据类型的表达式显式转换为另一种数据类型
cast( "abcdefg" as varchar(2000));

STUFF()函数：

--STUFF()函数用于删除指定长度的字符，并可以在制定的起点处插入另一组字符。

--select stuff(列名,开始位置,长度,替代字符串)

DISTINCT 关键字：

--这里可以理解为去重，具体可以百度

AS 关键字：

--取别名 

GROP BY ：

--用于分组

下面通过具体的例子来说明吧：    来自于：http://www.cnblogs.com/doubleliang/archive/2011/07/06/2098775.html

create table hobby(
hobbyId varchar(32),
hName varchar(32)
)

insert into hobby values(‘1‘,‘爬山‘);
insert into hobby values(‘2‘,‘游泳‘);
insert into hobby values(‘3‘,‘爬山‘);

--将查询结果根据行输出成XML格式
SELECT * FROM hobby FOR XML PATH;

--改变XML行节点的名称
SELECT * FROM hobby FOR XML PATH(‘MyHobby‘);

--改变XML列节点的名称
SELECT hobbyID as ‘MyCode‘,hName as ‘MyName‘ FROM hobby FOR XML PATH(‘MyHobby‘);

--构建我们喜欢的输出方式
SELECT ‘[ ‘+hName+‘ ]‘ FROM hobby FOR XML PATH(‘‘);

现在我对例子进行应用：

--应用
select distinct ‘,‘+hobbyId from hobby for xml path(‘‘);

-- for xml path(‘‘)  解决连接问题
select hName,(select distinct ‘+‘+hobbyId from hobby h where h.hName = hName for xml path(‘‘)) AS A 
from hobby 
group by hName;

-- stuff(String,1,1,‘‘) 去除开始的连接符  +1+2+3  --> 1+2+3
select hName,stuff((select distinct ‘+‘+hobbyId from hobby  for xml path(‘‘)),1,1,‘‘) AS A 
from hobby  
group by hName;

--需要查询结果
‘爬山‘ ‘1‘+‘3‘
‘游泳‘ ‘2‘

--方法一
select hName,stuff((select distinct ‘+‘+hobbyId from hobby where hName = B.hName for xml path(‘‘)),1,1,‘‘) AS A 
from hobby B 
group by hName;

--方法二
SELECT B.hName,LEFT(StuList,LEN(StuList)-1) as hobby FROM (
SELECT hName,
(SELECT hobbyId+‘,‘ FROM hobby 
  WHERE hName=A.hName 
  FOR XML PATH(‘‘)) AS StuList
FROM hobby A 
GROUP BY hName
) B 
结果：

第一次写笔记。。。。。2016.06.25

SQL基础知识--多行查询结果拼接

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