mysql：条件为条件的列的值为NULL时，选择查询 [英] mysql: select query when the column that has condition on has NULL value

问题描述

这是我的示例表：

CREATE TABLE test(
name VARCHAR(35),
age INT(3))

还有一些值：

insert into test values ('alex', 13);
insert into test values ('dan', 17);
insert into test (name) values ('pete');
insert into test (name) values ('david');

当我在条件为年龄的情况下使用SELECT查询时：

When I use SELECT query with the condition on column 'age':

select * from test where age!=13;

我得到的结果是：

+------+------+
| name | age  |
+------+------+
| dan  |   17 |
+------+------+

但是我想要所有具有年龄的记录！= 13，其中也包括具有年龄IS NULL的记录：

But I want all the record with age!=13 which includes the record with age IS NULL too:

+-------+------+
| name  | age  |
+-------+------+
| dan   |   17 |
| pete  | NULL |
| david | NULL |
+-------+------+

如何我能得到我想要的吗？

How can I get what I wants? Thanks for any responding.

推荐答案

SELECT * FROM test WHERE age!=13 OR age IS NULL;

请记住，NULL并不是真正的值，而更像是一个状态。 NULL = NULL 始终为false，因为 NULL！= NULL 。与 NULL 一起使用时，始终为false。因此，当它将 age！= 13 评估为 NULL！= 13 时，不显示该行是预期的行为

Keep in mind that NULL is not really a value but more like a state. NULL = NULL is always false as NULL != NULL. Whatever you use with NULL will always be false. So when it evaluates age != 13 to NULL != 13, it is the expected behavior to not show the row.

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