在 Oracle 中连接 SQL 查询的结果

Question

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|

我将如何在 Oracle 中进行查询以提供这样的字符串？我不需要将它编程为某种东西；我只是想要一种方法让该行出现在结果中，以便我可以将其复制并粘贴到 Word 文档中。

我的 Oracle 版本是 10.2.0.5。

Answer 1

-- 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

Answer 2

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

Answer 3

通常当我快速需要类似的东西并且我想在不使用 PL/SQL 的情况下继续使用 SQL 时，我会使用类似于以下 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

这是一个分层查询，使用“sys_connect_by_path”特殊函数，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，因此是第四个，然后获取最后一个子记录的完整路径（在这种情况下，记录 seq = 3 )，这将有效地串联所有“col”列

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

Answer 4

在 Oracle 10g 中，您不能使用出色的 listagg()。但是，还有许多其他string aggregation techniques。

没有特别需要所有复杂的东西。假设如下表

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

不支持的函数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，您也可以更改 Tom Kyte 的 stragg（也在上面的链接中），使其不使用替换功能。

Answer 5

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| 

SQLFiddle Example

Answer 6

类似以下的东西，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
    )  

Answer 7

使用 xmlagg 设法到达这里：使用 sql fiddle 中的 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|

* SQLFIDDLE DEMO

• Reference to read on XMLAGG