<?php
$today = date('Y-m-d');
$sql = "SELECT * FROM roombooking where enddate>='$today'" ;
$result1 = $db->query($sql);
if ($result1->num_rows > 0){
echo"
<div class=table-responsive border >
<table class=table>
<tr>
<th>Room Type</th>
<th>Room Number</th>
<th> Status </th>
<th> Check-in date </th>
<th> Check-Out date </th>
<th>Stay Time</th>
<th colspan='2'>Action </th>
</tr>";
// output data of each row
// output data of each row
while ($row = $result1->fetch_assoc())
{ echo "<tr>
<td>".$row["roomtype"]."</td>
<td>".$row["roomnumber"]."</td>
<td>".$row["bkg_status"]."</td>
<td>".$d1=$row["bkg_clientintime"]."</td>
<td>".$d2=$row["bkg_clientouttime"]."</td>
</tr>";
}
echo "</table></div>";
}else { echo "<br>"."<div class='alert alert-info col col-md-6 col-md-offset-2'> SORRY NO CHEK OUT FOUND FOR TODAY
</div> ";}
?>
//here from $d1 which is booking check-in-date and $d2 which is check-out-date
HELP REQUIRED // important checkout date time is 12.00 noon
HELP REQUIRED 2// FINDING TOTAL Number of days stay
any suggestion are greatly appreciated
tried here date_diff / it again gives hours or days but have to find based on
12.00 noon important checkout otherwise new day will be added how to achieve this part kindly guide
在结账时间是中午12点时,根据登记入住结算天数
- 写回答
- 好问题 0 提建议
- 追加酬金
- 关注问题
- 邀请回答
-
1条回答 默认 最新
- dongzong3053 2017-09-30 21:11关注
Here easiest way is calculate date different on mysql
SELECT *, DATE_DIFF(DATE(bkg_clientintime), DATE(bkg_clientouttime)) AS no_days FROM roombooking
本回答被题主选为最佳回答 , 对您是否有帮助呢?解决 无用评论 打赏 举报
悬赏问题
- ¥15 如何处理复杂数据表格的除法运算
- ¥15 如何用stc8h1k08的片子做485数据透传的功能?(关键词-串口)
- ¥15 有兄弟姐妹会用word插图功能制作类似citespace的图片吗?
- ¥200 uniapp长期运行卡死问题解决
- ¥15 请教:如何用postman调用本地虚拟机区块链接上的合约?
- ¥15 为什么使用javacv转封装rtsp为rtmp时出现如下问题:[h264 @ 000000004faf7500]no frame?
- ¥15 乘性高斯噪声在深度学习网络中的应用
- ¥15 关于docker部署flink集成hadoop的yarn,请教个问题 flink启动yarn-session.sh连不上hadoop,这个整了好几天一直不行,求帮忙看一下怎么解决
- ¥15 深度学习根据CNN网络模型,搭建BP模型并训练MNIST数据集
- ¥15 C++ 头文件/宏冲突问题解决