Question

基本上我想要添加field1，field2和field3中存储的数字，并且只显示此总数小于field4的记录...

例如：select * from table where field1 + field2 + field3 <= field 4

但那不起作用......任何帮助表示赞赏

这是我的实际SQL，因为你可以看到它更复杂：

SELECT
  *,
  SUM ( amountpaid , amountpaid2 , amountpaid3 ) AS total AND
  DATE_FORMAT(date_start, ' %d.%m.%Y') AS date_formatted
FROM 
  calendar_event 
WHERE 
  Date_start BETWEEN NOW() AND DATE_ADD(NOW(), INTERVAL 14 DAY) AND 
  total <= amount )

罗布

Answer 1

也许你的意思是：

SELECT
    *,
    amountpaid + amountpaid2 + amountpaid3 AS total,
    DATE_FORMAT(date_start, '%d.%m.%Y') AS date_formatted
FROM calendar_event
WHERE Date_start BETWEEN NOW() AND DATE_ADD(NOW(), INTERVAL 14 DAY)
AND amountpaid + amountpaid2 + amountpaid3 <= amount

假设您在amountpaid，amountpaid2和amountpaid3中分别付款，这将为您提供在接下来的14天内开始并且未支付过多的所有事件（即支付不足或支付的确实。）

Answer 2

您的查询

select * from table where field1 + field2 + field3 <= field4

应该完全正常。确保您要寻址的所有列实际存在，并且它们是数字（整数，浮点数，双精度等）。还要确保，如果你真的称之为，那个field4在“field”和“4”之间没有空格 - 你的问题之间有一个空格，比如“field 4”而不是“field4”。

或者，让我们知道您收到的错误消息。

Answer 3

Select
  *,
  Coalesce(amountpaid, 0) + Coalesce(amountpaid2, 0) + Coalesce(amountpaid3, 0) As total,
  DATE_FORMAT(date_start, ' %d.%m.%Y') As date_formatted
From
  calendar_event 
Where
  Date_start Between Now() And Date_Add(Now(), Interval 14 DAY) And
  Coalesce(amountpaid, 0) + Coalesce(amountpaid2, 0) + Coalesce(amountpaid3, 0) <= amount

http://sqlfiddle.com/#!2/35f6b/3

Mysql where query ... select * where field1 + field2 + field3＆lt; = field 4

3 个答案: