[英]Oracle Lag function with dynamic parameter
我有一個具體的問題。 我有一個包含無效值的表。 我需要將無效值(此處為0
)替換為大於0
的先前值。
困難在於,對我來說使用Update或插入是不合適的(Cursor和update會這樣做)。我唯一的方法是使用Select語句。
當我使用lag(col1, 1)
- 函數與大小寫時,我只得到一個具有正確值的列。
select col1, col2 realcol2,
(case
when col2 = 0 then
lag(col2,1,1) over (partition by col1 order by col1 )
else
col2
end ) col2,
col3 realcol3,
(case
when col3 = 0 then
lag(col3,1,1) over (partition by col1 order by col1 )
else
col3
end ) col3
from test_table
TEST_TABLE
內容:
---------------------------
Col1 | Col2 | Col3 | Col4
---------------------------
A | 0 | 1 | 5
B | 0 | 4 | 0
C | 2 | 0 | 0
D | 0 | 0 | 0
E | 3 | 5 | 0
F | 0 | 3 | 0
G | 0 | 3 | 1
A | 0 | 1 | 5
E | 3 | 5 | 0
預期的查詢結果:
---------------------------
Col1 | Col2 | Col3 | Col4
---------------------------
A | 0 | 1 | 5
B | 0 | 4 | 5
C | 2 | 4 | 5
D | 2 | 4 | 5
E | 3 | 5 | 5
F | 3 | 3 | 5
G | 3 | 3 | 1
A | 3 | 1 | 5
E | 3 | 5 | 5
我假設一個額外的列col0
包含一個明顯的數據排序條件,因為你的col1
示例數據沒有真正正確排序(重復, A
和E
尾隨值)。
我喜歡這些用途的MODEL
子句。 以下查詢產生預期結果:
WITH t(col0, col1, col2, col3, col4) AS (
SELECT 1, 'A', 0, 1, 5 FROM DUAL UNION ALL
SELECT 2, 'B', 0, 4, 0 FROM DUAL UNION ALL
SELECT 3, 'C', 2, 0, 0 FROM DUAL UNION ALL
SELECT 4, 'D', 0, 0, 0 FROM DUAL UNION ALL
SELECT 5, 'E', 3, 5, 0 FROM DUAL UNION ALL
SELECT 6, 'F', 0, 3, 0 FROM DUAL UNION ALL
SELECT 7, 'G', 0, 3, 1 FROM DUAL UNION ALL
SELECT 8, 'A', 0, 1, 5 FROM DUAL UNION ALL
SELECT 9, 'E', 3, 5, 0 FROM DUAL
)
SELECT * FROM t
MODEL
DIMENSION BY (row_number() OVER (ORDER BY col0) rn)
MEASURES (col1, col2, col3, col4)
RULES (
col2[any] = DECODE(col2[cv(rn)], 0, NVL(col2[cv(rn) - 1], 0), col2[cv(rn)]),
col3[any] = DECODE(col3[cv(rn)], 0, NVL(col3[cv(rn) - 1], 0), col3[cv(rn)]),
col4[any] = DECODE(col4[cv(rn)], 0, NVL(col4[cv(rn) - 1], 0), col4[cv(rn)])
)
結果:
RN COL1 COL2 COL3 COL4
1 A 0 1 5
2 B 0 4 5
3 C 2 4 5
4 D 2 4 5
5 E 3 5 5
6 F 3 3 5
7 G 3 3 1
8 A 3 1 5
9 E 3 5 5
雖然上面看起來很酷(或可怕,取決於你的觀點),你當然應該更喜歡使用基於窗口函數的appraoch,通過nop77svk(使用LAST_VALUE() IGNORE NULLS
NULLS )或MT0(使用LAG() IGNORE NULLS
)的其他優雅答案公開。 LAG() IGNORE NULLS
) 。 我在這篇博客文章中更詳細地解釋了這些答案 。
假設您希望按照原始數據順序 (無論可能是什么)獲得先前的值,那么您的查詢可能如下所示:
with preserve_the_order$ as (
select X.*,
rownum as original_order$
from test_table X
)
select X.col1,
nvl(last_value(case when col2 > 0 then col2 end) ignore nulls over (order by original_order$ rows between unbounded preceding and current row), col2) as col2,
nvl(last_value(case when col3 > 0 then col3 end) ignore nulls over (order by original_order$ rows between unbounded preceding and current row), col3) as col3,
nvl(last_value(case when col4 > 0 then col4 end) ignore nulls over (order by original_order$ rows between unbounded preceding and current row), col4) as col4
from preserve_the_order$ X
order by original_order$
;
結果:
COL1 COL2 COL3 COL4
---- ---------- ---------- ----------
A 0 1 5
B 0 4 5
C 2 4 5
D 2 4 5
E 3 5 5
F 3 3 5
G 3 3 1
A 0 1 5
E 3 5 5
SELECT col1,
CASE col2 WHEN 0 THEN NVL( LAG( CASE col2 WHEN 0 THEN NULL ELSE col2 END ) IGNORE NULLS OVER ( ORDER BY NULL ), 0 ) ELSE col2 END AS col2,
CASE col3 WHEN 0 THEN NVL( LAG( CASE col3 WHEN 0 THEN NULL ELSE col3 END ) IGNORE NULLS OVER ( ORDER BY NULL ), 0 ) ELSE col3 END AS col3,
CASE col4 WHEN 0 THEN NVL( LAG( CASE col4 WHEN 0 THEN NULL ELSE col4 END ) IGNORE NULLS OVER ( ORDER BY NULL ), 0 ) ELSE col4 END AS col4
FROM table_name;
結果 :
COL1 COL2 COL3 COL4
---- ---------- ---------- ----------
A 0 1 5
B 0 4 5
C 2 4 5
D 2 4 5
E 3 5 5
F 3 3 5
G 3 3 1
A 3 1 5
E 3 5 5
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