根据时间范围，我有一个mySQL表来指示商店开放或关闭的天气：
shift table
--------------
shift_id
day_of_week (1-7 : Monday-Sunday)
open_time (time, default NULL)
close_time (time, default NULL)
type (1 || 2)
字段类型说明： 1 表示同一天内的第一个时间范围， 2 表示时间范围是同一天内的第二个时间范围。 例如，商店在13:00 - 15:00和18:00 - 01:00开放。
我使用以下方法获取当天：
$jd = cal_to_jd(CAL_GREGORIAN,date("m"),date("d"),date("Y"));
$day = jddayofweek($jd, 0);
switch($day){
case 0:
$curDay = 7;
break;
default:
$curDay = $day;
break;
}
return $curDay;
我使用以下方法获取当前时间：
$timeString = "%H:%i:%s";
$curTime = mdate($timeString, time());
复杂的例子如下：
当天：星期一
当前时间：02：00
周日时间：15：00-18：00和21:00 - 02:30
周一时间表：08：30 - 15:30。
显然，这种转变是 OPEN ，但是基于周日的时间范围，而不是周一的当天。
在给定日期的给定时间内，查询我的表格（或在需要时更改我的表格）或100％准确的方法来定义每次轮班是开放还是关闭的最佳方式是什么？
答案 0 :(得分：0)
function shop_open(){
//Get current day
$day = $this->_get_current_day();
//Get current time
$timeString = "%H:%i:%s";
$ordertime = mdate($timeString, time());
//Predefine ability to order to false
$canOrder = false;
$getShift = $this->db->query("select * from `shop_hours` where `day_of_week` = $day AND !ISNULL(`open_time`) ORDER BY `type`");
$day_num_rows = $getShift->num_rows();
switch($day_num_rows){
case 0: //No records
$canOrder = false;
break;
case 1: //Only 1 timeframe
foreach($getShift->result() as $row){
$open_time = $row->open_time;
$close_time = $row->close_time;
}
if($open_time <= $close_time){//Timeframe is within the same day [no overlap]
if($open_time < $ordertime && $close_time > $ordertime){//Current time is within timeframe
$canOrder = true;
}else{
$canOrder = false;
}
}else if($open_time > $close_time){//There s your overlap [timframe "catches" the next day]
if($open_time < $ordertime && '23:59' >= $ordertime){//Check till midnight first
$canOrder = true;
}else{//Check in the previous day
if($day == 1){ $previousDay = 7; }else{ $previousDay = $day-1; }
$getPrvDayShift = $this->db->query("select * from `shop_hours` where `day_of_week` = $previousDay ORDER BY `type`");
$prv_day_num_rows = $getPrvDayShift->num_rows();
switch($prv_day_num_rows){
case 0://No records
$canOrder = false;
break;
case 1://One record found
foreach($getPrvDayShift->result() as $row2){
$prv_close_time = $row2->close_time;
}
if($prv_close_time > $ordertime){ //So if the previous day close time is greater than current, we are good to go
$canOrder = true;
}else{
$canOrder = false;
}
break;
case 2: //2nd record found
foreach($getPrvDayShift->result() as $row2){
if($row2->type == 2){
$prv_close_time = $row2->close_time;
}
}
if($prv_close_time > $ordertime){ //Same shit, different timeframe within the same day
$canOrder = true;
}else{
$canOrder = false;
}
break;
}
}
}
break;
case 2: //Ok now we have to check the case of 2 timeframes together
for($i=1; $i<3; $i++){
foreach($getShift->result() as $row){
if($row->type == $i){
$open_time = $row->open_time;
$close_time = $row->close_time;
}
}
if($open_time <= $close_time){//Again, same day [no overlap]
if($open_time < $ordertime && $close_time > $ordertime){//Good to go we are within timeframe
$canOrder = true;
}else{
$canOrder = false;
}
}else if($open_time > $close_time){//There s your overlap
//Check till midnight
if($open_time < $ordertime && '23:59' >= $ordertime){ $canOrder = true; }
}
}
if(!$canOrder){
//Now we should check the previous day also if $canOrder is still negative
if($day == 1){ $previousDay = 7; }else{ $previousDay = $day-1; }
$getPrvDayShift = $this->db->query("select * from `shop_hours` where `day_of_week` = $previousDay ORDER BY `type`");
$prv_day_num_rows = $getPrvDayShift->num_rows();
switch($prv_day_num_rows){
case 0:
$canOrder = false;
break;
case 1:
foreach($getPrvDayShift->result() as $row2){
$prv_close_time = $row2->close_time;
}
if($prv_close_time > $ordertime){ $canOrder = true; }else{ $canOrder = false; }
break;
case 2:
foreach($getPrvDayShift->result() as $row2){
if($row2->type == 2){
$prv_close_time = $row2->close_time;
}
}
if($prv_close_time > $ordertime){ $canOrder = true; }else{ $canOrder = false; }
break;
}
}
break;
}
return $canOrder;
}
$ this-＆gt; _get_current_day（）就在我的问题中。
我想必须有更优雅和安全的解决方案。
答案 1 :(得分：0)
为了区分延伸到第二天的时间范围与不支持的时间范围，时间范围不能超过24小时。因此，这些时间范围的结束时间必须小于或等于开放时间。
此查询应该在给定时间找到开放班次：
SELECT shift_id
FROM shifts
WHERE :curDay = day_of_week
AND :curTime >= open_time -- today
AND (:curTime <= close_time OR close_time <= open_time) -- shift may end tomorrow
OR :curDay = IF(day_of_week=1,7,day_of_week-1) -- check yesterday
AND :curTime <= close_time AND close_time <= open_time -- shift started yesterday
如果可能的话，我会更改表格，将跨越午夜的班次分成两班，一班结束于24:00，另一班则从第二天的00:00开始。
答案 2 :(得分：0)
这就是我的想法：
CREATE TABLE Shifts (
shift_id INT AUTO_INCREMENT PRIMARY KEY,
start_day CHAR(9) NOT NULL,
start_time TIME NOT NULL,
end_day CHAR(9) NOT NULL,
end_time TIME NOT NULL
) ENGINE = InnoDB;
现在我想到了，我真的不确定你是否需要end_day。但是，使用：
INSERT INTO Shifts (
start_day,
start_time,
end_day,
end_time
) VALUES
( LOWER('Sunday'), '15:00:00', LOWER('Sunday'), '18:30:00' ),
( LOWER('Sunday'), '21:00:00', LOWER('Monday'), '26:30:00' ),
( LOWER('Monday'), '12:00:00', LOWER('Monday'), '18:00:00' ),
( LOWER('Tuesday'), '10:00:00', LOWER('Tuesday'), '20:45:00' ),
( LOWER('Wednesday'), '16:00:00', LOWER('Wednesday'), '19:30:00' ),
( LOWER('Thursday'), '10:00:00', LOWER('Thursday'), '17:00:00' ),
( LOWER('Thursday'), '19:00:00', LOWER('Friday'), '25:30:00' ),
( LOWER('Friday'), '16:00:00', LOWER('Saturday'), '24:30:00' ),
( LOWER('Saturday'), '15:00:00', LOWER('Saturday'), '20:00:00' ),
( LOWER('Saturday'), '18:30:00', LOWER('Sunday'), '27:00:00' )
然后您可以使用以下内容：
SELECT DAYNAME(NOW()), start_day
FROM Shifts
WHERE (start_day = LOWER(DAYNAME(NOW()))
AND start_time < CURTIME()
AND end_time > CURTIME())
OR (start_day = LOWER(DAYNAME(DATE_SUB(NOW(), INTERVAL 1 DAY)))
AND start_time < ADDTIME('24:00:00', CURTIME())
AND end_time > ADDTIME('24:00:00', CURTIME()))
你可以在这里测试一下：
http://sqlfiddle.com/#!2/41a26/34
尝试使用不同的测试值：
SELECT DAYNAME(NOW()), start_day
FROM Shifts
WHERE (start_day = LOWER('Friday')
AND start_time < '01:12:35'
AND end_time > '01:12:35')
OR (start_day = LOWER('Thursday')
AND start_time < ADDTIME('24:00:00', '01:12:35')
AND end_time > ADDTIME('24:00:00', '01:12:35'))