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我有兩個表。我想創建一個視圖。我寫了一個查詢,用正確的名稱替換了第二個表的值。甲骨文有很多空值
S_ID | STATION_NAME
------ | ------
1 | PACKAGER1
2 | PACKAGER2
3 | PACKAGER3
4 | PACKAGER4
5 | PACKAGER5
6 | PACKAGER6
7 | PACKAGER7
C_ID | STATION_ID | TO_STATION_ID
------ | -----------| -------------
1 | 1 | 7
2 | 2 | 7
3 | 3 | 7
4 | 4 | 7
5 | 5 | 7
6 | 6 | 7
7 | 7 | 1
7 | 7 | 2
7 | 7 | 3
7 | 7 | 4
7 | 7 | 5
7 | 7 | 6
SELECT CC.STATION_ID AS NUM,
CASE WHEN CC.STATION_ID = S.S_ID THEN S.STATION_NAME END
AS "FROM STATION"
FROM CONNECTION CC,
STATIONS S
替換工作正常,但創建了很多其他空值的單元格。全部12 * 7 = 82。如果兩個值不相等,我如何跳過這種情況。
NUM | FROM_STATION
------ | ------
1 | PACKAGER1
2 | null
3 | null
4 | null
5 | null
6 | null
7 | null
我希望得到這樣的結果:
NUM | FROM_STATION | TO_STATION
------ | ------------ | ----------
1 | PACKAGER1 | Packager7
2 | PACKAGER2 | Packager7
3 | PACKAGER3 | Packager7
4 | PACKAGER4 | Packager7
5 | PACKAGER5 | Packager7
6 | PACKAGER6 | Packager7
7 | PACKAGER7 | Packager1
8 | PACKAGER7 | Packager2
9 | PACKAGER7 | Packager3
10 | PACKAGER7 | Packager4
11 | PACKAGER7 | Packager5
12 | PACKAGER7 | Packager6
希望你所需要的戈登·利諾夫回答。請參考這個鏈接,https://www.techonthenet.com/oracle/joins.php希望這會給你一些想法。 – vipin