Question

我想在SQLite中创建一个视图，其中一行中的字段取决于前一行中字段的值。我可以使用LAG分析函数在Oracle中执行此操作，但不确定如何在SQLite中进行此操作。

例如，如果我的表格如下：

ITEM        DAY           PRICE
apple       2011-01-07    1.25
orange      2011-01-02    1.00
apple       2011-01-01    1.00
orange      2011-01-03    2.00
apple       2011-01-08    1.00
apple       2011-01-10    1.50

我希望我的观点看起来像WHERE item = 'apple'：

DAY           PRICE    CHANGE
2011-01-01    1.00     (null)
2011-01-07    1.25     0.25
2011-01-08    2.00     0.75
2011-01-10    1.50     -0.50

编辑：

相当于我正在寻找的查询会在Oracle中看起来像（没试过这个，但我认为这是正确的）：

SELECT day, price, 
       price - LAG( price, 1 ) OVER ( ORDER BY day ) AS change
  FROM mytable
 WHERE item = 'apple'

Answer 1

它与另一个相同，但只使用字段而不是rowid。这完全符合您的要求：


CREATE TABLE Prices (
    day DATE,
    price FLOAT
);

INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '+1 day'), 0.5);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '+0 day'), 1);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '-1 day'), 2);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '-2 day'), 7);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '-3 day'), 8);
INSERT INTO Prices(day, price) VALUES(date('now', 'localtime', '-4 day'), 10);

SELECT p1.day, p1.price, p1.price-p2.price 
FROM
    Prices p1, Prices p2,
    (SELECT t2.day AS day1, MAX(t1.day) AS day2 
    FROM Prices t1, Prices t2
    WHERE t1.day < t2.day
    GROUP BY t2.day) AS prev
WHERE p1.day=prev.day1
    AND p2.day=prev.day2

如果您要添加WHERE item='apple'位，请将其添加到WHERE个子句中。

Answer 2

这应该适用于每个item（在SQLite上测试）：

SELECT 
    day
    ,price
    ,price - (SELECT t2.price 
              FROM mytable t2 
              WHERE 
                  t2.item = t1.item AND 
                  t2.day < t1.day      
              ORDER BY t2.day DESC
              LIMIT 1
             ) AS change
FROM mytable t1

这假设day和item之间的组合是唯一的。它的工作方式是将所有值小于给定day，排序降序，然后排序LIMIT只是第一个值，模拟一个LAG函数。

对于LEAD行为，只需将<转为>，将DESC转为ASC。

Answer 3

Oracle等效项是正确的。从SQLite 3.25.0开始，您可以本地使用LAG：

WITH mytable(ITEM,DAY,PRICE) AS (
    VALUES
    ('apple',  CAST('20110107' AS DATE),    1.25),
    ('orange', CAST('20110102' AS DATE),    1.00),
    ('apple',  CAST('20110101' AS DATE),    1.00),
    ('orange', CAST('20110103' AS DATE),    2.00),
    ('apple',  CAST('20110108' AS DATE),    2.00),
    ('apple',  CAST('20110110' AS DATE),    1.50)
)
SELECT day, price, price-LAG(price) OVER (ORDER BY day) AS change
FROM mytable
WHERE item = 'apple'
ORDER BY DAY;

Answer 4

假设你没有删除它将起作用：


SELECT t2.DAY, t2.price, t2.price-t1.price 
FROM TABLENAME t1, TABLENAME t2 
WHERE t1.rowid=t2.rowid-1

这是有效的，因为即使你没有在CREATE语句中指定它，每一行都有自己的rowid。

如果删除，则变为：


SELECT t2.day, t2.price, t2.price-t1.price 
FROM 
     (SELECT l1.day, l1.price, 
          (SELECT COUNT(*) 
          FROM TABLENAME l2 
          WHERE l2.rowid < l1.rowid) AS count
      FROM TABLENAME l1) AS t1,
     (SELECT l1.day, l1.price, 
          (SELECT COUNT(*) 
          FROM TABLENAME l2 
          WHERE l2.rowid < l1.rowid) AS count
      FROM TABLENAME l1) AS t2
WHERE t1.count=t2.count-1

这可以假设rowid总是在增加。

创建一个SQLite视图，其中行取决于前一行

4 个答案: