我在表格中有这样的数据
NAME PRICE
A 2
B 3
C 5
D 9
E 5
我想在一行中显示所有值;例如:
A,2|B,3|C,5|D,9|E,5|
如何在Oracle中创建一个类似字符串的查询?我不需要把它编程成某种东西;我只想要一种方法让这条线出现在结果中,这样我就可以复制它并将其粘贴到word文档中。
我的Oracle版本是10.2.0.5。
答案 0 :(得分:9)
- Oracle 10g -
SELECT deptno, WM_CONCAT(ename) AS employees
FROM scott.emp
GROUP BY deptno;
Output:
10 CLARK,MILLER,KING
20 SMITH,FORD,ADAMS,SCOTT,JONES
30 ALLEN,JAMES,TURNER,BLAKE,MARTIN,WARD
答案 1 :(得分:5)
我知道这有点晚了但试试这个:
SELECT LISTAGG(CONCAT(CONCAT(NAME,','),PRICE),'|') WITHIN GROUP (ORDER BY NAME) AS CONCATDATA
FROM your_table
答案 2 :(得分:3)
通常当我需要这样的东西很快并且我想在不使用PL / SQL的情况下继续使用SQL时,我会使用类似于下面的hack的东西:
select sys_connect_by_path(col, ', ') as concat
from
(
select 'E' as col, 1 as seq from dual
union
select 'F', 2 from dual
union
select 'G', 3 from dual
)
where seq = 3
start with seq = 1
connect by prior seq+1 = seq
这是一个使用“sys_connect_by_path”特殊功能的分层查询,该功能旨在获取从父级到子级的“路径”。
我们正在做的是模拟seq = 1的记录是seq = 2的记录的父级,然后是第四个,然后得到最后一个孩子的完整路径(在这种情况下,seq = 3的记录) ),这将有效地成为所有“col”列的串联
适应您的情况:
select sys_connect_by_path(to_clob(col), '|') as concat
from
(
select name || ',' || price as col, rownum as seq, max(rownum) over (partition by 1) as max_seq
from
(
/* Simulating your table */
select 'A' as name, 2 as price from dual
union
select 'B' as name, 3 as price from dual
union
select 'C' as name, 5 as price from dual
union
select 'D' as name, 9 as price from dual
union
select 'E' as name, 5 as price from dual
)
)
where seq = max_seq
start with seq = 1
connect by prior seq+1 = seq
结果是:|A,2|B,3|C,5|D,9|E,5
答案 3 :(得分:2)
当你在Oracle 10g中时,你不能使用优秀的listagg()
。但是,还有许多其他string aggregation techniques。
并不特别需要所有复杂的东西。假设下表
create table a ( NAME varchar2(1), PRICE number);
insert all
into a values ('A', 2)
into a values ('B', 3)
into a values ('C', 5)
into a values ('D', 9)
into a values ('E', 5)
select * from dual
不支持的功能wm_concat
应该足够了:
select replace(replace(wm_concat (name || '#' || price), ',', '|'), '#', ',')
from a;
REPLACE(REPLACE(WM_CONCAT(NAME||'#'||PRICE),',','|'),'#',',')
--------------------------------------------------------------------------------
A,2|B,3|C,5|D,9|E,5
但是,您也可以在上面的链接中更改Tom Kyte的stragg
,以便在没有替换功能的情况下执行此操作。
答案 4 :(得分:2)
这是另一种方法,使用model
子句:
-- sample of data from your question
with t1(NAME1, PRICE) as(
select 'A', 2 from dual union all
select 'B', 3 from dual union all
select 'C', 5 from dual union all
select 'D', 9 from dual union all
select 'E', 5 from dual
) -- the query
select Res
from (select name1
, price
, rn
, res
from t1
model
dimension by (row_number() over(order by name1) rn)
measures (name1, price, cast(null as varchar2(101)) as res)
(res[rn] order by rn desc = name1[cv()] || ',' || price[cv()] || '|' || res[cv() + 1])
)
where rn = 1
结果:
RES
----------------------
A,2|B,3|C,5|D,9|E,5|
答案 5 :(得分:1)
使用xmlagg管理到这里:使用来自sql小提琴的oracle 11G。
数据表:
COL1 COL2 COL3
1 0 0
1 1 1
2 0 0
3 0 0
3 1 0
SELECT
RTRIM(REPLACE(REPLACE(
XMLAgg(XMLElement("x", col1,',', col2, col3)
ORDER BY col1), '<x>'), '</x>', '|')) AS COLS
FROM ab
;
结果:
COLS
1,00| 3,00| 2,00| 1,11| 3,10|
答案 6 :(得分:0)
如下所示,效率极低且未经测试。
create function foo returning varchar2 as
(
declare bar varchar2(8000) --arbitrary number
CURSOR cur IS
SELECT name,price
from my_table
LOOP
FETCH cur INTO r;
EXIT WHEN cur%NOTFOUND;
bar:= r.name|| ',' ||r.price || '|'
END LOOP;
dbms_output.put_line(bar);
return bar
)