我希望使用CONNECT BY LEVEL根据列值复制行。
这是我使用的代码:
SELECT
tbl1.*,
(tbl1.START_WEEK - tbl1.END_WEEK) wks_inbtwn
FROM
My_SQL_table tbl1 INNER JOIN (SELECT rownum repeat FROM dual CONNECT BY LEVEL <= tbl1.END_WEEK ) tbl2
ON tbl2.Repeat > tbl1.START_WEEK
我不断收到错误消息:
SQL Error [904] [42000]: ORA-00904: "tbl1"."START_WEEK": invalid identifier
我的表格如下:
+------------+------------+----------+
| Site_NUM | start_week | end_week |
+------------+------------+----------+
| France | 50 | 52 |
| Germany | 41 | 43 |
| USA | 12 | 13 |
+------------+------------+----------+
我想要的结果如下:
+----------+---------+
| Site_NUM | Week_no |
+----------+---------+
| France | 51 |
| France | 52 |
| Germany | 42 |
| Germany | 43 |
| USA | 13 |
+----------+---------+
任何帮助将不胜感激,谢谢。
答案 0 :(得分:1)
Oracle设置:
CREATE TABLE My_SQL_table ( Site_NUM, start_week, end_week ) AS
SELECT 'France', 50, 52 FROM DUAL UNION ALL
SELECT 'Germany', 41, 43 FROM DUAL UNION ALL
SELECT 'USA', 12, 13 FROM DUAL;
查询:使用CONNECT BY
SELECT site_num,
COLUMN_VALUE wks_inbtwn
FROM My_SQL_table tbl1
CROSS JOIN
TABLE(
CAST(
MULTISET(
SELECT tbl1.START_WEEK + LEVEL
FROM DUAL
CONNECT BY tbl1.START_WEEK + LEVEL <= tbl1.END_WEEK
)
AS SYS.ODCINUMBERLIST
)
)
输出:
SITE_NUM | WKS_INBTWN :------- | ---------: France | 51 France | 52 Germany | 42 Germany | 43 USA | 13
查询2 :使用递归子查询分解子句
WITH rsqfc ( site_num, start_week, end_week ) AS (
SELECT site_num, start_week + 1, end_week
FROM my_sql_table
UNION ALL
SELECT site_num, start_week + 1, end_week
FROM rsqfc
WHERE start_week < end_week
)
SELECT site_num, start_week AS wks_inbtwn
FROM rsqfc
ORDER BY site_num, wks_inbtwn
输出:
SITE_NUM | WKS_INBTWN :------- | ---------: France | 51 France | 52 Germany | 42 Germany | 43 USA | 13
db <>提琴here