【发布时间】:2016-09-20 21:01:15
【问题描述】:
我试图根据我的数据库中的打开和关闭时间来显示商店是打开还是关闭。如果它是开放的,则显示当天的开放和关闭时间,如果它是关闭的,则回显关闭。我目前的问题是,即使商店计划营业(尝试 1)或根本没有回显任何内容(尝试 2),我的查询也会回显已关闭。
我的数据库中关闭的商店表示为 00:00。 任何建议或指导将不胜感激,因为我正在自学并且已经停滞不前。
数据库
CREATE TABLE `Opening_hrs` (
`OH_ID` bigint(255) NOT NULL AUTO_INCREMENT,
`Restaurant_ID` bigint(255) NOT NULL,
`Day_of_week` int(11) NOT NULL,
`Open_time` time NOT NULL,
`Closing_time` time NOT NULL,
PRIMARY KEY (`OH_ID`),
KEY `Restaurant_ID` (`Restaurant_ID`)
) ENGINE=InnoDB AUTO_INCREMENT=8 DEFAULT CHARSET=utf8
这是我的第一次尝试
date_default_timezone_set("Europe/London");
$output_ohr = '';
$ohrs = mysqli_query($dbc, "SELECT * FROM Opening_hrs
WHERE Restaurant_ID='$rest_id' AND Day_of_week = DATE_FORMAT(NOW(), '%w')
AND CURTIME() BETWEEN Open_time AND Closing_time");
echo var_dump($ohrs);
$count_ohrs = mysqli_num_rows($ohrs);
if ($count_ohrs === 0) {
$output_ohr = '<b> Closed</b>';
} else {
$i = 1;
}while ($row_ohr = mysqli_fetch_array($ohrs )) {
$o_time = $row_ohr['Open_time'];
$c_time = $row_ohr['Closing_time'];
$output_ohr = $output_ohr . '<p>Open</p>' .
'<p>' .$o_time. ' - ' .$c_time. '</p>'
;
$i++;
}
我的第二次尝试
date_default_timezone_set("Europe/London");
$closed= strtotime("00:00am today GMT");
$output_ohr = '';
$ohrs = mysqli_query($dbc, "SELECT * FROM Opening_hrs
WHERE Restaurant_ID='$rest_id' AND Day_of_week = DATE_FORMAT(NOW(), '%w')
AND CURTIME() BETWEEN Open_time AND Closing_time");
echo var_dump($ohrs);
$i = 1;
while ($row_ohr = mysqli_fetch_array($ohrs )) {
$o_time = $row_ohr['Open_time'];
$c_time = $row_ohr['Closing_time'];
if($o_time === $closed){
$output_ohr = '<p>closed</p>';
}else{
$output_ohr = $output_ohr . '<p>Open</p>' .
'<p>' .$o_time. ' - ' .$c_time. '</p>'
;
$i++;
}
}
【问题讨论】: