SQLite - 获得最接近的价值 [英] SQLite - getting closest value

问题描述

我有 SQLite 数据库，其中有double"类型的特定列.我想获得该列值最接近指定值的行.

I have SQLite database and I have in it certain column of type "double". I want to get a row that has in this column value closest to a specified one.

例如，在我的表中，我有:

For example, in my table I have:

id: 1; value: 47
id: 2; value: 56
id: 3; value: 51

我想得到一个值最接近 50 的行.所以我想接收 id: 3 (value = 51).

And I want to get a row that has its value closest to 50. So I want to receive id: 3 (value = 51).

我怎样才能实现这个目标?

How can I achieve this goal?

谢谢.

推荐答案

这应该有效:

SELECT * FROM table
ORDER BY ABS(? - value)
LIMIT 1

其中 ? 代表您要与之比较的值.

Where ? represents the value you want to compare against.

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