[英]MySQL - Sort on a calculated value based on two dates
我有一个需要在MySQL查询中解决的以下问题:
Fields
info - textfield
date1 - a date field
date2 - a date field
offset1 - a text field with a number in the first two positions, example "10-High"
offset2 - a text field with a number in the first two positions, example "10-High"
我需要根据当前日期(今天)按计算的“sortvalue”对记录进行排序:
If today<date1 then sortvalue=offset1*10+offset2*5-1000
else if today>=date2 then sortvalue=offset1*10+offset2*5+1000
else sortvalue=offset1*10+offset2*5
我对连接等基本SQL有很好的理解,但是我甚至不确定它是否可能......如果它有助于我可以使用单一的公式,给出与IF相同的效果....即。 在date1 =低值之前,在date2 =高值之后...
Rgds PM
如果将自定义排序值添加到选择列表,则可以直接在ORDER BY子句中使用它。 就像是:
SELECT field1, field2, fieldn,
IF(today < @date1, offset1 * 10 + offset2 * 5 - 1000,
IF(today >= @date2, offset1 * 10 + offset2 * 5 + 1000,
offset1 * 10 + offset2 * 5
)
) AS sortvalue
ORDER BY sortvalue;
采用:
...
FROM FIELDS t
ORDER BY CASE
WHEN NOW() < t.date1 THEN CAST(LEFT(offset1, 2) AS INT) * 10 + CAST(LEFT(offset2, 2) AS INT) * 5 - 1000
WHEN NOW() >= t.date2 THEN CAST(LEFT(offset1, 2) AS INT) * 10 + CAST(LEFT(offset2, 2) * 5 + 1000
ELSE offset1 * 10 + offset2 * 5
END
你想要一个SQL“CASE”子句。 请查看http://dev.mysql.com/doc/refman/5.5/en/control-flow-functions.html#operator_case
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