我嘗試了許多oracle分析函數來實現我的需求,但無法實現。 這裏是一個簡單的例子:甲骨文獲得超過訂單組的第一價值
WITH W1 AS (
SELECT 100 key1, 0 key2, 0 key3, 'open' status, 'date1' date_column FROM DUAL UNION ALL
SELECT 100 key1, 0 key2, 1 key3, 'open' status, 'date2' date_column FROM DUAL UNION ALL
SELECT 100 key1, 0 key2, 2 key3, 'close' status, 'date3' date_column FROM DUAL UNION ALL
SELECT 100 key1, 0 key2, 3 key3, 'close' status, 'date4' date_column FROM DUAL UNION ALL
SELECT 100 key1, 0 key2, 4 key3, 'close' status, 'date5' date_column FROM DUAL UNION ALL
SELECT 100 key1, 0 key2, 5 key3, 'open' status, 'date6' date_column FROM DUAL UNION ALL
SELECT 100 key1, 0 key2, 6 key3, 'open' status, 'date7' date_column FROM DUAL)
SELECT W1.*,
CASE WHEN LAG(status,1) OVER(PARTITION BY key1,key2 ORDER BY key3) <> status THEN date_column
ELSE FIRST_VALUE(date_column) OVER (PARTITION BY key1,key2,status ORDER BY key3 RANGE BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) END as DESIRED
FROM W1
ORDER BY 1,2,3
我想,當狀態改變克服組有序第一個值。 我只需要在DESIRED列上的最後一行應該是「date6」
你有什麼建議嗎? Oracle 11gr2
謝謝。
你能告訴你想要的輸出? –
只有查詢結果的最後一行需要是「date6」 – Deniz