将列与可能（或不）为空的值进行比较的最简单方法

Question

我有一个准备好的 SQLite 语句...

SELECT * FROM a WHERE col1 = $someValue

但是当someValue是null时查询returns没有行，所以我写了这个...

SELECT *
FROM a
WHERE (
    col1 = $someValue OR
    (
        $someValue IS NULL AND
        col1 IS NULL
    )
)

无论 someValue 是否为 null，此查询都能完美运行，但它非常冗长。 是否有更简单或更简洁的方法来实现同样的目标？

Answer 1

如果你要使用 postgres 有 is distinct from 我想他们有它是因为很多人和你有同感:

SELECT * FROM a WHERE col1 is not distinct from $someValue

使用 sqlite 应该可以这样写：

SELECT * FROM a WHERE col1 is $someValue OR col1 = $someValue

这可能是最短的了。

Answer 2

来自Operators, and Parse-Affecting Attributes：

The IS and IS NOT operators work like = and != except when one or both of the operands are NULL. In this case, if both operands are NULL, then the IS operator evaluates to 1 (true) and the IS NOT operator evaluates to 0 (false). If one operand is NULL and the other is not, then the IS operator evaluates to 0 (false) and the IS NOT operator is 1 (true). It is not possible for an IS or IS NOT expression to evaluate to NULL.

您也可以使用运算符 IS 来比较非空值：

SELECT * FROM a WHERE col1 IS $someValue;

Answer 3

不如 forpas 的答案优雅，但另一个值得一提的选择：

SELECT * FROM a WHERE ifnull(col1, 'x') = ifnull($someValue, 'x')

传递给 ifnull 函数的第二个参数，在本例中为 x，是数据库中不存在的值，col1 和 $someValue 是 varchar列。

将列与可能（或不）为空的值进行比较的最简单方法

Simplest way to compare column against a value that may (or not) be null

sql

sqlite

node-sqlite3