SQLite：Select n 到 n-y 范围内的最小值

Question

我有一个由 ID 和客户标识的订单列表。我需要知道从该客户放置的索引 n 的订单开始的索引 n-9 的订单的 id。

例如：

SELECT OrderID 
FROM Orders 
WHERE CustomerID = "1808317576" AND OrderID < 29 
ORDER BY OrderID DESC LIMIT 9

上面的查询returns这些记录：

订单号
28
26
23
20
19
18
17
16
13

为了解决我的问题，我尝试这样做：

SELECT OrderID 
FROM Orders 
WHERE CustomerID = "1808317576" AND OrderID < 29 
ORDER BY OrderID DESC LIMIT 1

我需要：

订单号
13

但是 returns 这个:

订单号
28

然后我尝试从 ORDER BY 中删除 DESC:

SELECT OrderID 
FROM Orders 
WHERE CustomerID = "1808317576" AND OrderID < 29 
ORDER BY OrderID LIMIT 1

但是 returns 这个:

订单号
1

怎样才能得到13？！

Answer 1

只需在相反（升序）方向子查询您当前的查询和顺序，限制为 1:

SELECT OrderID
FROM
(
    SELECT OrderID
    FROM Orders
    WHERE CustomerID = '1808317576' AND OrderID < 29
    ORDER BY OrderID DESC
    LIMIT 9
) t
ORDER BY OrderID
LIMIT 1;

Answer 2

添加 OFFSET 8 以跳过返回的前 8 行并设置 LIMIT 1:

SELECT OrderID 
FROM Orders 
WHERE CustomerID = '1808317576' AND OrderID < 29 
ORDER BY OrderID DESC 
LIMIT 1 OFFSET 8

或更简单：

SELECT OrderID 
FROM Orders 
WHERE CustomerID = '1808317576' AND OrderID < 29 
ORDER BY OrderID DESC 
LIMIT 8,1

SQLite：Select n 到 n-y 范围内的最小值

SQLite: Select minimum value in a range from n to n-y

sqlite

sql-order-by

sql-limit