我已经创造了某种依赖关系"包含我们夜间进程的整个依赖关系的表。
表格如下:
GRAND_MODEL | WAIT_4_MODEL_NAME
test test1
test test2
test test3
test2 test3
test3 test4
test4 test5
此表表示 - > test
需要等待test1
,test2
,test3
才能完成,但还需要等待test4
和test5
,因为test3/4
等待他们。
test1
并不等待任何事情,test2
等待test3
,因此test4
也因此test5
。
所以结果应该是这样的:
FIRST_MODEL | SECOND_MODEL | THIRD_MODEL | FORTH_MODEL | FIFTH_MODEL | SIXTH_MODEL
test5 test4 test3 test2 test NULL
test5 test4 test3 test NULL NULL
test4 test3 test2 test NULL NULL
.................................
我尝试过的事情:
SELECT distinct prior wait_4_model_name as first_m,
wait_4_model_name as second_m,
grand_model as third_m
from (SELECT distinct grand_model, wait_4_model_name
FROM DEL_SAGI_FOR_HIERARCHY)
connect by NOCYCLE prior grand_model = wait_4_model_name
但这只会生成层次结构的第一级。
提前致谢。
编辑:请注意,可能存在相反的依赖关系,test
等待test1
而test1
等待test
(每个模型都很大) ,因此模型的一部分可以等待另一个模型的一部分)
答案 0 :(得分:2)
Oracle安装程序:
CREATE TABLE table_name ( GRAND_MODEL, WAIT_4_MODEL_NAME ) AS
SELECT 'test', 'test1' FROM DUAL UNION ALL
SELECT 'test', 'test2' FROM DUAL UNION ALL
SELECT 'test', 'test3' FROM DUAL UNION ALL
SELECT 'test2', 'test3' FROM DUAL UNION ALL
SELECT 'test3', 'test4' FROM DUAL UNION ALL
SELECT 'test4', 'test5' FROM DUAL;
<强>查询强>:
SELECT REGEXP_SUBSTR( tests, '[^|]+', 1, 1 ) AS first_model,
REGEXP_SUBSTR( tests, '[^|]+', 1, 2 ) AS second_model,
REGEXP_SUBSTR( tests, '[^|]+', 1, 3 ) AS third_model,
REGEXP_SUBSTR( tests, '[^|]+', 1, 4 ) AS fourth_model,
REGEXP_SUBSTR( tests, '[^|]+', 1, 5 ) AS fifth_model,
REGEXP_SUBSTR( tests, '[^|]+', 1, 6 ) AS sixth_model,
REGEXP_SUBSTR( tests, '[^|]+', 1, 7 ) AS seventh_model,
REGEXP_SUBSTR( tests, '[^|]+', 1, 8 ) AS eighth_model,
REGEXP_SUBSTR( tests, '[^|]+', 1, 9 ) AS ninth_model
FROM (
SELECT SYS_CONNECT_BY_PATH( wait_4_model_name, '|' ) || '|' || grand_model AS tests
FROM table_name
CONNECT BY PRIOR grand_model = wait_4_model_name
);
<强>输出强>:
FIRST_MODEL SECOND_MODEL THIRD_MODEL FOURTH_MODEL FIFTH_MODEL SIXTH_MODEL SEVENTH_MODEL EIGHTH_MODEL NINTH_MODEL
----------- ------------ ----------- ------------ ----------- ----------- ------------- ------------ -----------
test1 test
test2 test
test3 test
test3 test2
test3 test2 test
test4 test3
test4 test3 test
test4 test3 test2
test4 test3 test2 test
test5 test4
test5 test4 test3
test5 test4 test3 test
test5 test4 test3 test2
test5 test4 test3 test2 test
答案 1 :(得分:1)
这是一个不需要字符串连接和重新拆分的解决方案:
select a.wait_4_model_name model1,
a.grand_model model2,
b.grand_model model3,
c.grand_model model4,
d.grand_model model5,
e.grand_model model6
from DEL_SAGI_FOR_HIERARCHY a
left join DEL_SAGI_FOR_HIERARCHY b ON a.grand_model = b.wait_4_model_name
left join DEL_SAGI_FOR_HIERARCHY c ON b.grand_model = c.wait_4_model_name
left join DEL_SAGI_FOR_HIERARCHY d ON c.grand_model = d.wait_4_model_name
left join DEL_SAGI_FOR_HIERARCHY e ON d.grand_model = e.wait_4_model_name
where a.wait_4_model_name not in (
select grand_model from DEL_SAGI_FOR_HIERARCHY)
order by 1, 2, 3, 4, 5, 6
样本数据的输出是:
+--------+--------+--------+--------+--------+--------+
| MODEL1 | MODEL2 | MODEL3 | MODEL4 | MODEL5 | MODEL6 |
+--------+--------+--------+--------+--------+--------+
| test1 | test | - | - | - | - |
| test5 | test4 | test3 | test | - | - |
| test5 | test4 | test3 | test2 | test | - |
+--------+--------+--------+--------+--------+--------+
请注意,您的示例数据没有完全没有任何依赖关系的模型实例,即不需要等待其他模型,也不需要任何模型才能启动。