【发布时间】：2018-06-09 01:11:07
【问题描述】：
我正在制作预订系统，更新数据时出现问题。我做了这样的表updateMy_ReservationView.php。
this is an image of SelectMy_ReservationView.php 由于此编辑器的错误消息，我插入更多代码时出错。
<?php
include "connection.php";
$id=$_GET['reservation_id'];
$sql = "select reservation.*, customer.*, car_type.*, datediff(return_time,
rent) as total_day, (datediff(return_time, rent) * price ) AS total_price
FROM
reservation, customer, car_type
WHERE reservation.car_type_id=car_type.car_type_id AND
reservation.customer_id=customer.customer_id and reservation_id='$id' order
by reservation_id ";
$result = mysqli_query($conn, $sql);
$row = mysqli_fetch_assoc($result);
$sql_car = "SELECT car_type.* from car_type";
$result_car = mysqli_query($conn, $sql_car);
?>
<h3><b>Update Reservation</b></h3><br>
<form method = "post" action = "?page=updateMy_ReservationDo">
<table class="table table-striped table-sm"
style="width:500px; height:200px;">
<tr>
<td>Customer Name</td>
<td>
<?php echo" $row[customer_name]";?>
<input type = "hidden" name="reservation_id" value="
<?php echo"$row[reservation_id]";?>">
</td>
</tr>
<tr>
<td>Old car type</td>
<td>
<?php echo" $row[car_type]";?>
</td>
</tr>
<tr>
<td>New Car Type (Price USD)</td>
<td>
<select name = "car_type">
<?php
while($row_car = mysqli_fetch_assoc($result_car)) {
?>
<option value="<?php echo"$row_car[car_type_id]";?>">
<?php echo"$row_car[car_type] ($row_car[price])";?></option>
<?php
}
?>
</select>
</td>
</tr>
<tr>
<td>Old Rent</td>
<td><?php echo "$row[rent]"; ?></td>
</tr>
<tr>
<td>Rent</td>
<td><input type="text" name="rent" id="rent"
maxlength="25" size="25"/>
<img src="images_date/cal.gif" alt=""
onclick="javascript:NewCssCal('rent','yyyyMMdd','arrow',false,'24',false)"
style="cursor:pointer"/></td>
</tr>
<tr>
<td>Old Return</td>
<td><?php echo "$row[return_time]"; ?></td>
</tr>
<tr>
<td>Return</td>
<td><input type="text" name="return_time"
id="return_time" maxlength="25" size="25"/>
<img src="images_date/cal.gif" alt=""
onclick="javascript:NewCssCal('return_time','yyyyMMdd'
'arrow',false,'24',false)" style="cursor:pointer"/></td>
</tr>
<tr>
<td>Old Pickup Station</td>
<td><?php echo "$row[car_station]"; ?></td>
</tr>
<tr>
<td>Pickup Station</td>
<td>
<select name = "car_station">
<option value="Yeouido">Yeouido</option>
<option value="Shinchon">Shinchon</option>
<option value="Jongro">Jongro</option>
<option value="Seoul Station">Seoul
Station</option>
<option value="Gangnam">Gangnam</option>
<option value="Geondae">Geondae</option>
</select></td>
</tr>
<tr>
<td> </td>
<td><input type="reset" value="Reset"> <input name = "add" type = "submit" value = "Update Reservation">
</td>
</tr>
</table>
我做了更新函数文件 updateMy_ReservationDo.php 如下所示。
include "connection.php";
$reservation_id=$_POST['reservation_id'];
$car_type=$_POST['car_type_id'];
$rent=$_POST['rent'];
$return_time=$_POST['return_time'];
$car_station=$_POST['car_station'];
$sql = "update reservation set car_type='$car_type_id',rent='$rent',
return_time='$return_time' and car_station='$car_station' where
reservation_id=$reservation_id ";
if (mysqli_query($conn, $sql)) {
echo "Reservation is updated successfully<br>";
echo "<p><p><a href=?page=selectMy_reservationView><button type=button>Show
all reservation</button></a>";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
然后出现这样的错误消息：
Notice: Undefined index: car_type_id in C:\xampp\htdocs\rentcar\updateMy_ReservationDo.php on line 5
Notice: Undefined variable: car_type_id in C:\xampp\htdocs\rentcar\updateMy_ReservationDo.php on line 10
Error: update reservation set car_type='',rent='2018-05-31', return_time='2018-06-01' and car_station='Shinchon' where reservation_id=17
Unknown column 'car_type' in 'field list'
我应该修改什么？
【问题讨论】：
-
根据我的经验，价格会随着季节的变化而波动，所以我无法想象这样的场景可以简单地通过将天数乘以一个数字来计算。
-
感谢您的评论。这只是简单的本科生项目，所以我们没有考虑。价格仅由汽车计算（例如，奥迪 a5、现代索纳塔等）我看到了您关于 selectMy_reservationView 的最后一条评论，但由于代码太多，我无法上传它......通过堆栈编辑器的错误消息。你有没有错误上传的想法？如果需要，我可以把它作为图片发布吗？