Question

我想使用Php和mysql每半小时显示一次商店的可用性状态，

为此，我尝试使用以下代码为我创建时隙

$duration="30";
    $start="10:00AM";
    $end="07:00PM";

    $start = new DateTime($start);
        $end = new DateTime($end);
        $start_time = $start->format('H:i');
        $end_time = $end->format('H:i');
        $i=0;
        while(strtotime($start_time) <= strtotime($end_time)){
            $start = $start_time;
            $end = date('H:i',strtotime('+'.$duration.' minutes',strtotime($start_time)));
            $start_time = date('H:i',strtotime('+'.$duration.' minutes',strtotime($start_time)));
            $i++;
            if(strtotime($start_time) <= strtotime($end_time)){
                $time[$i]['start'] = $start;
                $time[$i]['end'] = $end;
            }
        }
        print_R($time);

上面的代码显示了以下结果（正确创建时隙）

Array
(
    [1] => Array
        (
            [start] => 10:00
            [end] => 10:30
        )

    [2] => Array
        (
            [start] => 10:30
            [end] => 11:00
        )
    ...// and so on

我想要这样的结果

Array
(
    [1] => Array
        (
            [start] => 10:00
            [end] => 10:30
            [status] => availiable
        )

    [2] => Array
        (
            [start] => 10:30
            [end] => 11:00
            [status] => booked
        )

这是我在phpmyadmin中的“预订”表

id       shop_id     date        booking_start_time    booking_close_time
1        3           4-7-2019        10:00                 11:00

我该怎么做？预先感谢

Answer 1

您需要按以下方式比较数据库中的时隙：

while(strtotime($start_time) <= strtotime($end_time)){
    $start = $start_time;
    $end = date('H:i',strtotime('+'.$duration.' minutes',strtotime($start_time)));
    $start_time = date('H:i',strtotime('+'.$duration.' minutes',strtotime($start_time)));
    $i++;
    if(strtotime($start_time) <= strtotime($end_time)){
        $time[$i]['start'] = $start;
        $time[$i]['end'] = $end;
    }
    //Here you need to write query and fetch data.
    $todayDate = date('d-m-Y'); //Please check date format. It should be similar as your database date field format.
    //please use data binding instead of contacting variable.
    $selectQuery = "select `id` from `booking` where date = "'.$todayDate.'" and 
        (( `booking_start_time` >= "'.$start.'" AND `booking_start_time` <= "'.$start.'" ) || 
        (`booking_close_time` >= "'.$end.'" AND `booking_close_time` <= "'.$end.'")) ";

    // After, you need to exeucte this query and need to check query output. if it has records, then you need to show booked else available. as below
    $result = mysqli_query($con, $selectQuery);
    if ($result->num_rows) {
        $time[$i]['status'] = 'booked';
    } else {
        $time[$i]['status'] = 'availiable';
    }
}
print_R($time);

希望它对您有帮助。

Answer 2

这是您查询的最佳解决方案。

select count(*) as aggregate from `bookings` where (`booking_date` = '$booking_date') and (( start_time >= '$start_time' AND end_time <= '$end_time' ) || (end_time >= '$start_time' AND start_time <= '$end_time'))

用php中的时隙检查时间的可用性

2 个答案: