Question

这是我的电视指南脚本的一部分：

       //Connect to the database
   mysql_connect("localhost","root","PASSWORD");
   //Select DB
   mysql_select_db("mytvguide");

   //Select only results for today and future
   $result = mysql_query("SELECT programme, channel, episode, airdate, expiration, setreminder FROM mediumonair where airdate >= now()");

剧集出现了，所以那里没有问题。但是，它正在让数据库找到问题所在的数据。

如果我为今天播放的节目添加记录，则应显示：

Medium   showing on TV4   8:30pm    "Episode"    Set Reminder
Medium  showing on TV4  May 18th - 6:25pm  "Episode 2"  Set Reminder
Medium  showing on TV4  May 18th - 10:25pm  "Episode 3"  Set Reminder
Medium  showing on TV4  May 19th - 7:30pm  "Episode 3"  Set Reminder
Medium  showing on TV4  May 20th - 1:25am  "Episode 3"  Set Reminder
Medium  showing on TV4  May 20th - 6:25pm  "Episode 4"  Set Reminder

但是这反过来显示：

    Medium  showing on TV4  May 18th - 6:25pm  "Episode 2"  Set Reminder
Medium  showing on TV4  May 18th - 10:25pm  "Episode 3"  Set Reminder
Medium  showing on TV4  May 19th - 7:30pm  "Episode 3"  Set Reminder
Medium  showing on TV4  May 20th - 1:25am  "Episode 3"  Set Reminder
Medium  showing on TV4  May 20th - 6:25pm  "Episode 4"  Set Reminder

我几乎让SQL工作;只是不确定这里有什么正确的代码，以避免第二个错误显示 - 因为记录（表示正在播出的节目）目前似乎不起作用。

请有人帮我这个吗？

由于

Answer 1

我不知道，但我猜你在日期和时间中存储日期和时间的方式有些奇怪。尝试运行这样的查询，这样您就可以进行与数据库要求相同的比较：

select now(), airdate from mediumonair where episode = 'Episode'

电视指南脚本 - 获取当前日期节目

1 个答案: