2016-05-23 73 views
0

我试图根据打开和关闭时间在我的数据库中显示店铺是否打开或关闭。如果是打开显示当天的开启和关闭时间,则关闭回显关闭。我目前的问题是,即使店铺计划开放(尝试1)或根本没有回应任何东西(尝试2),我的查询仍然是封闭的。店铺开门和关门时间,打开时显示关闭

我DB中的封闭商店被表示为00:00。 任何建议或指导将不胜感激,因为我正在教我自己,并已停下来。

DB

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++; 
     } 
    }  

回答

0

在你的第二次尝试,你只能选择开放商店。如果商店关闭,查询将不会返回任何行,因此while块不会运行一次。

使用下面的代码,看看商店是打开还是关闭:

$stmt = mysqli_prepare($dbc, "SELECT COUNT(*) FROM Opening_hrs 
WHERE Restaurant_ID=? AND Day_of_week = DATE_FORMAT(NOW(), '%w') 
AND CURTIME() BETWEEN Open_time AND Closing_time"); 

$stmt->bind_param('i', $rest_id); 
$stmt->execute(); 
$stmt->bind_result($result); 
$stmt->fetch(); 

if ($result) { 
    echo 'Open'; 
} else { 
    echo 'Closed'; 
} 

(注意的prepared statements防止可能的SQL注入的使用)

+0

谢谢您的帮助和时间,但是这也行不通 – Blue

+0

@蓝色你能更具体吗? – ShiraNai7

+0

即使餐厅开放,它也会关闭。 – Blue