mysql哪里有条件问题

Question

可以使用像这样的WHERE;

$no = $_GET['no'];
$query = "SELECT * FROM `numbers` WHERE `myno` - $no = 4";
...

如果没有，我怎样才能获得如下例所示的行？

编辑： 更清楚，

$today = date("Y-m-d");
 $query = "SELECT * FROM `dates` WHERE date - $today = 7";

我想找到日期字段为7天前的行

THX

Answer 1

如果该字段是日期字段

$query = "SELECT * FROM `numbers` WHERE `myno` = DATE_SUB(`myno`, INTERVAL 7 DAY)"

编辑：或者我会优先考虑

$query = "SELECT * FROM `numbers` WHERE DATEDIFF(NOW(), `myno`) = 7"

Take a look at the date functions

Answer 2

您可以在WHERE子句中执行各种表达式，但要小心：如果您这样做

SELECT * FROM numbers WHERE myno - ? = 4

它将无法使用任何索引，而是：

SELECT * FROM numbers WHERE myno = 4 + ?

将能够在myno上使用索引。

Answer 3

你可以这样做：

$query = "SELECT * FROM `numbers` WHERE `myno` = 4  + $no"

Answer 4

$ weekAgo = date（“Y-m-d”，strtotime（“ - 1周”））;

$ query =“SELECT * FROM numbers WHERE myno ='$ weekAgo';