我完全疯了，我自己都想不出来，我已经读了我能读到的每一篇文章，但遗憾的是什么都没有。所以我希望有人能帮忙。
我有一张桌子，里面放着表演。它有四个时间字段show_start（HH:MM AM/PM）、show_end（HH:M AM/PM），show_start_actual（HH:MM:SS）和show_end_actual（HH:MM:SS）。此外，我还有7个以一周中的几天（周日、周一等）命名的专栏。一周中的几天是1或0，以注释节目是否在某一天。actual列保持24小时时间格式，其他时间为varchar字段。
我正试着撤下目前正在播放的节目。时间很基本（比如早上6点到10点等）我在PHP和MySQL中尝试过，但我所做的一切都不起作用。我相信有一个简单的解决办法。
这是我当前在PHP中比较时间的格式副本，但我认为如果我能在mySQL中实际将其提取出来，效率会更高
型号
function on_air()
{
$current_time = $this->convert_to_hour(date('g:i a',now()));
$current_day_of_week = strtolower(date('l',now()));
$mySQL = $this->db->query("SELECT * FROM on_air_now");
if($mySQL->num_rows() > 0 ){
$mySQL = $mySQL->result();
foreach($mySQL as $row){
eval('$db_day_of_week = $row->'.$current_day_of_week.';');
$show_start = $this->convert_to_hour($row->show_start);
$show_end = $this->convert_to_hour($row->show_end);
if($current_time > $show_start && $current_time < $show_end && $db_day_of_week == 1){
// If true then go at the show tables to get info...
}
}
}else{
return FALSE;
}
}
function convert_to_hour($str){
$temp_arr = explode(":",$str);
$arr_cnt = count($temp_arr)-1;
$hour = $temp_arr[0];
$min = explode(" ",$temp_arr[$arr_cnt]);
$ampm = $min[1];
if($ampm == 'pm' || $ampm == 'PM'){
if($hour !== '12'){
$hour = $hour+12;
}
}else{
if($hour == '12'){
$hour = 0;
}
}
$new_time = $hour;
return $new_time;
}
更新数据：
id,"datetime","show_name","sub_name","show_start","show_end","show_start_actual","show_end_actual","day_of_week",sunday,monday,tuesday,wednesday,thursday,friday,saturday,id_personalities,active,"date_added"
1,"2011-06-29 11:33:46","DJ","DJ","06:00 am","10:00 am","06:00:00","10:00:00",NULL,NULL,1,1,1,1,1,1,0,1,NULL
2,"2011-06-29 11:33:46","DJ","DJ","12:00 pm","03:00 pm","12:00:00","03:00:00",NULL,NULL,1,1,1,1,1,NULL,0,1,NULL
3,"2011-06-29 11:33:46","DJ","DJ","09:00 am","12:00 pm","09:00:00","12:00:00",NULL,NULL,0,0,0,0,0,NULL,NULL,1,NULL
4,"2011-06-29 11:35:17","DJ","DJ","03:00 pm","06:00 pm","15:00:00","18:00:00",NULL,NULL,1,1,1,1,1,0,NULL,1,NULL
5,"2011-06-29 11:35:39","Scott Stevens","Scott Stevens","06:00 pm","09:00 pm","18:00:00","21:00:00",NULL,NULL,0,0,0,0,0,NULL,NULL,1,"2011-06-21 6:15:57"
6,"2011-06-29 11:37:03","DJ","DJ","09:00 pm","12:00 am","21:00:00","00:00:00",NULL,NULL,0,0,0,0,0,NULL,NULL,1,"2011-06-21 6:18:10"
7,"2011-06-29 11:37:17","DJ","DJ","12:00 am","06:00 am","00:00:00","06:00:00",NULL,NULL,1,1,1,1,1,NULL,NULL,NULL,NULL
这是表中的CSV转储。
首先，您可以从简化数据库设计开始。以下是一个建议的替代方案。
CREATE TABLE `show` (
`id` INT UNSIGNED NOT NULL AUTO_INCREMENT,
`show_name` VARCHAR(64) NOT NULL,
`sub_name` VARCHAR(64) NOT NULL,
`show_start` DATETIME NOT NULL,
`show_end` DATETIME NOT NULL,
/* OTHER COLUMNS */
`datecreated` TIMESTAMP NOT NULL DEFAULT CURRENT_TIMESTAMP,
`dateupdated` TIMESTAMP,
PRIMARY KEY (`id`)
) ENGINE=INNODB DEFAULT CHARSET=utf8;
除非节目在一周内重复多次，否则你不需要本周的七栏。只需使用MySQL函数从show_start本身获取即可。
一旦您完成了这项工作，使用MySQL或PHP获取所有当前的节目都是微不足道的。例如，要使用MySQL获取今天的所有节目，查询将是
SELECT * FROM shows
WHERE DATE(NOW()) = DATE(`show_start`)
在上面的查询中也有很多DATE的替代项。