分層查詢

Question

我創建了某種“依賴項”表，其中包含了我們夜間流程的所有依賴項。

該表如下所示：

GRAND_MODEL | WAIT_4_MODEL_NAME
test            test1
test            test2
test            test3
test2           test3
test3           test4
test4           test5

該表意味着-> test為了完成需要等待test1 ， test2 ， test3 ，但它也需要等待test4和test5 ，因為test3/4等待它們。 test1不等待任何東西， test2等待test3 ，因此也等待test4 ，因此也等待test5 。

因此結果應如下所示：

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
.................................

我嘗試過的

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

但這只會生成層次結構的第一級。

提前致謝。

編輯：請注意，可以有相反的依賴關系， test等待test1和test1等待test （每個模型都很大，因此該模型的一部分有可能等待另一個模型的一部分）

Answer 1

Oracle安裝程序 ：

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;

查詢 ：

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
);

輸出 ：

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                                                           

Answer 2

這是不需要字符串連接和重新分割的解決方案：

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

樣本數據的輸出為：

+--------+--------+--------+--------+--------+--------+
| MODEL1 | MODEL2 | MODEL3 | MODEL4 | MODEL5 | MODEL6 |
+--------+--------+--------+--------+--------+--------+
| test1  | test   |    -   |   -    |   -    |   -    |
| test5  | test4  | test3  | test   |   -    |   -    |
| test5  | test4  | test3  | test2  | test   |   -    |
+--------+--------+--------+--------+--------+--------+

請注意，您的樣本數據沒有完全沒有任何依賴關系的模型實例，即不需要等待其他模型，也不需要任何模型在啟動之前就不需要。