问题:oracle LISTAGG 连接字符串;结果:Oracle 连接字符串的方法

时间:2021-09-19 19:43:01

Oracle 连接字符串的方法


方法一:wmsys.wm_concat(column)

 

    介绍:其函数在Oracle 10g推出,在10g版本中,返回字符串类型,在11g版本中返回clob类型。括号里面的参数是列,而且可以是多个列的集合,也就是说在括号里面可以*地用‘||’合并字符串。如下面的例子:

Select u_id, wmsys.wm_concat(goods || '(' || num || '斤)' ) goods_sum   from shopping   group by u_id

方法二:listagg (column,[,]) within group (order by ) [over (partition by  )]

 

         介绍:其函数在Oracle 11g 版本中推出,对分组后的数据按照一定的排序进行字符串连接。其中,“[,]”表示字符串连接的分隔符,如果选择使用[over (partition by )]则会使其变成分析函数;

 

方法三:sys_connect_by_path(column,<分隔符>)

 

        介绍:其函数在Oracle 9i 版本中推出,用来合并链路的字符串。注意的是其一定要和connect by子句合用!第一个参数是形成树形式的字段,第二个参数是父级和其子级分隔显示用的分隔符。

 

下面是上面几种方法的实例(在Oracle 11g版本中运行正确):

 

 

实例:

方法一:用listagg(,',') within group()

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WITH temp1 AS(

select 'a' as username,1 as deptid from dual union all

select 'b',1 from dual union all

select 'c',1 from dual union all

select 'd',2 from dual union all

select 'e',2 from dual

),

temp2 AS(

select 1 as deptid,'部门1' as deptname from dual union all

select 2 ,'部门2' from dual

)

select p.deptid,

listagg(t.username,',') within group (order by t.username) as username,

p.deptname

from temp1 t,temp2 p

where t.deptid=p.deptid

group by p.deptid,p.deptname

order by p.deptid

-------------------------

deptid userName deptName

1 a,b,c 部门1

2 d,e 部门2



方法二:用wm_concat()

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select p.deptid,

wm_concat(t.username) as username,

p.deptname

from temp1 t,temp2 p

where t.deptid=p.deptid

group by p.deptid,p.deptname

order by p.deptid



方法三:用CONNECT BY

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select deptid,

ltrim(max(sys_connect_by_path(username,',')),','),

deptname

from (

select p.deptid,t.username,p.deptname,

row_number()over(partition by t.deptid order by t.username) as ar

from temp1 t,temp2 p

where t.deptid=p.deptid

)

start with ar=1

connect by prior ar=ar-1

and username=prior username

group by deptid,deptname

order by deptid

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