我有一个有趣的问题,至少我是这么认为的。 所以,我有一个表(codes_table)与上面的结构(它是一个树菜单) 树由两个整数和一个点图案(总是)定义,如下所示:
COD_MENU NAME
01. Biggest Node
01.01. Sun of 01.
01.01.01. Sun of 01.01.
01.01.01.01. Sun of 01.01.01.
01.01.01.02. Sun of 01.01.01.
01.01.01.03. Sun of 01.01.01.
01.01.01.04. Sun of 01.01.01.
01.01.01.05. Sun of 01.01.01.
01.02. Sun of 01.
01.03. Sun of 01.
01.03.01. Sun of 01.03.
(etc...)
我需要的是为具有数字结构的新表生成插入脚本 带有foreing键的父亲的ID将实际的COD_MENU作为名称放在 像这样的新结构:
ID FATHER_ID NAME
1 NULL 01.
2 1 01.01.
3 2 01.01.01.
4 3 01.01.01.01.
5 3 01.01.01.02.
6 3 01.01.01.03.
7 3 01.01.01.04.
8 3 01.01.01.05.
9 1 01.02.
10 1 01.03.
11 10 01.03.01.
我已经在带有递归过程的plsql块代码中执行了此操作。 以上是我的真实代码供参考。
declare
auxId integer;
procedure findNodes( pCod varchar2, pCurrId in out integer, pFatherId in integer ) is
ctHasSuns integer;
father varchar2(20);
idFhtTmp integer;
begin
idFhtTmp := pFatherId;
if idFhtTmp is null then
father := 'null';
else
father := idFhtTmp;
end if;
ctHasSuns := 0;
SELECT count(cod_menu) into ctHasSuns FROM codes_table where SUBSTR (cod_menu, 1,LENGTH (cod_menu) - 3) = pCod and rownum < 10;
if (ctHasSuns > 0 or pCurrId = 1) then
dbms_output.put_line ('insert into newtable ( id, idfather, menu ) values ( '||pCurrId||','||father||', '''||pCod||''' );');
idFhtTmp := pCurrId;
for cHasSuns in ( SELECT cod_menu FROM codes_table where SUBSTR (cod_menu, 1,LENGTH (cod_menu) - 3) = pCod and rownum < 10 order by cod_menu) loop
pCurrId := pCurrId + 1;
findNodes( cHasSuns.cod_menu, pCurrId, idFhtTmp );
end loop;
else
dbms_output.put_line ('insert into newtable ( id, idfather, menu ) values ( '||pCurrId||','||father||', '''||pCod||''' );');
end if;
end;
begin
auxId := 1;
findNodes( '01.', auxId, null );
end;
我正在尝试做的是使用CONNECT BY / START WITH / ROWNUM在单个查询中完成 如有必要,LEVEL语句和usign联接使用相同的表。我尝试过很多东西但不能 出来了一个解决方案。这只是为了好奇。
我能做的最接近的事情就是上面的选择,但我只有父亲的身份证而不是太阳队
WITH q AS (
select rownum id, father from (
select a.noh father, count(*)
from (select cod_menu noh, substr( cod_menu, 1, length(cod_menu)-3) as nofather from codes_table) a,
(select cod_menu noh, substr( cod_menu, 1, length(cod_menu)-3) as nofather from codes_table) b
where a.noh = b.nofather
group by a.noh
having count(*) > 0
order by a.noh)
),
b as (
SELECT cod_menu filho, father
FROM (select cod_menu, substr( cod_menu, 1, length(cod_menu)-3) as father from codes_table)
START WITH father is null
CONNECT BY prior cod_menu = father
ORDER BY cod_menu
)
SELECT * FROM q, b where q.father = b.father;
感谢您的时间和帮助。
答案 0 :(得分:1)
尝试以下解决方案:
CREATE TABLE codes_table (
cod_menu VARCHAR2(40),
name VARCHAR2(40)
);
INSERT INTO codes_table VALUES ('01.', 'Biggest Node');
INSERT INTO codes_table VALUES ('01.01.', 'Sun of 01.');
INSERT INTO codes_table VALUES ('01.01.01.', 'Sun of 01.01.');
INSERT INTO codes_table VALUES ('01.01.01.01.', 'Sun of 01.01.01.');
INSERT INTO codes_table VALUES ('01.01.01.02.', 'Sun of 01.01.01.');
INSERT INTO codes_table VALUES ('01.01.01.03.', 'Sun of 01.01.01.');
INSERT INTO codes_table VALUES ('01.01.01.04.', 'Sun of 01.01.01.');
INSERT INTO codes_table VALUES ('01.01.01.05.', 'Sun of 01.01.01.');
INSERT INTO codes_table VALUES ('01.02.', 'Sun of 01.');
INSERT INTO codes_table VALUES ('01.03.', 'Sun of 01.');
INSERT INTO codes_table VALUES ('01.03.01.', 'Sun of 01.03.');
COMMIT;
SELECT
id,
PRIOR id AS father_id,
cod_menu AS name
FROM (
SELECT
cod_menu,
name,
row_number() OVER (ORDER BY cod_menu) AS id
FROM
codes_table
)
START WITH cod_menu = '01.'
CONNECT BY SUBSTR(cod_menu, 1, LENGTH(cod_menu) - 3) = PRIOR cod_menu
;
检查SQLFiddle:Fiddle with it