选择今天、本周、本月的记录 php mysql

问题描述

我想这很简单，但无法弄清楚.我正在尝试制作几页 - 其中包含从我的 mysql db 表中选择的今天、本周和本月的结果.日期是在使用 date('Y-m-d H:i:s'); 创建记录时输入的.这是我到目前为止所拥有的:

I imagine this is pretty simple, but can't figure it out. I'm trying to make a few pages - one which will contain results selected from my mysql db's table for today, this week, and this month. The dates are entered when the record is created with date('Y-m-d H:i:s');. Here's what I have so far:

day where date>(date-(60*60*24))

day where date>(date-(60*60*24))

 "SELECT * FROM jokes WHERE date>(date-(60*60*24)) ORDER BY score DESC"

日期所在的星期>(日期-(60*60*24*7))

week where date>(date-(60*60*24*7))

 "SELECT * FROM jokes WHERE date>(date-(60*60*24*7)) ORDER BY score DESC"

月(30天)其中日期>(日期-(60*60*24*30))

month (30 days) where date>(date-(60*60*24*30))

 "SELECT * FROM jokes WHERE date>(date-(60*60*24*30)) ORDER BY score DESC"

任何想法将不胜感激.谢谢！

Any ideas would be much appreciated. Thanks!

推荐答案

假设您的日期列是实际的 MySQL 日期列:

Assuming your date column is an actual MySQL date column:

SELECT * FROM jokes WHERE date > DATE_SUB(NOW(), INTERVAL 1 DAY) ORDER BY score DESC;        
SELECT * FROM jokes WHERE date > DATE_SUB(NOW(), INTERVAL 1 WEEK) ORDER BY score DESC;
SELECT * FROM jokes WHERE date > DATE_SUB(NOW(), INTERVAL 1 MONTH) ORDER BY score DESC;

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