从今天选择记录，本周本月，php mysql [英] Select records from today, this week, this month php mysql

问题描述

我想象这很简单，但不能弄清楚。我试图做几页 - 一个将包含今天，本周和本月从我的mysql数据库表中选择的结果。使用 date（'Y-m-d H：i：s'）; 创建记录时，将输入日期。这是我到目前为止：

日期>（日期 - （60 * 60 * 24））

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

周日期>（日期 - （60 * 60 * 24 * 7））

 code>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））

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

任何想法都将不胜感激。谢谢！

解决方案

假设你的日期列是一个实际的MySQL日期列：

  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; 
  

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))

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

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

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

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

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

Any ideas would be much appreciated. Thanks!

解决方案

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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