Question

我是PHP的新手，我一直在尝试从两个不同的表中插入值＃34;用户＆＃34;并通过下拉列表将＃34; avail＆＃34;带到第三个空表＆＃34;预订＆＃34;。另外，我还希望手动插入2个数据＆＃34; Start_Time＆＃34;和＆＃34; End_Time＆＃34;进入第三个表。我设法用提交按钮调出表单。

这是我的代码（我在顶部省略了部分代码）：

<html>
<head>
<title> Booking form for Carpark </title>
<meta http-equiv="Content-Type" content ="text/html; charset=iso-8859-1">
</head>
<body>

<form name="form1" action="bookingssql.php" method="post">

<label type='text'>UserID:</label>
<select name ='UserID'>
<?php 
$conn = new mysqli('localhost', 'root','','carpark'); 
$result1 = $conn->query("select UserID from user");
  while($row =$result1->fetch_assoc())
    { ?>
      <option value="<? php echo $row['UserID']; ?>">
      <?php echo $row['UserID']; ?>
      </option>
      <?php
    } ?>

  </select> 
  <br>

<label type='text'> Development:</label>
<select name ='Development'>
<?php 
$conn1 = new mysqli('localhost', 'root','','carpark'); 
$result = $conn1->query("select Development from avail");
  while($row =$result->fetch_assoc())
    { ?>
      <option value="<? php echo $row['Development']; ?>">
      <?php echo $row['Development']; ?>
      </option>
      <?php
      } ?>

  </select>
   <br>

<table border = 3, cellpadding=2,cellspacing=1>
<tr>
<th>Start Time     </th>
<th>End Time     </th>
<th> </th>
</tr>
<td><input type=text name=Start_Time>
<td><input type=text name=End_Time>
<input type=submit value = Book>
</form>

</table>        
</body>
</html>

但是，单击按钮后无法插入值，我不知道为什么。这是剩下的代码，我认为这里出错：

<?php
$con = new mysqli('localhost','root','','carpark');

$ID = $_POST["UserID"];
$Dev = $_POST["Development"];
$Start=$_POST["Start_Time"];
$End= $_POST["End_Time"];
//Insert Query
$sql = "INSERT INTO bookings (UserID,Development,Start_Time,End_Time) VALUES('$ID','$Dev','$Start','$End')";
$result=mysqli_query($con,$sql);

if(mysqli_query($con,$sql))
{
    $message = "Booking Made!";
    echo "<script type='text/javascript'>alert('$message');</script>";
    header("refresh:1; url=bookings.php");
}
else
{
    echo "Not Booked";
}



?>

希望我能带上一双新的眼睛来帮助我发现错误。非常感谢提前！

Answer 1

为该选项命名，请尝试使用：

import Amplify from 'aws-amplify-react-native'

和/或：

<option name="UserID" value="<?php echo $row['UserID']; ?>

将两个不同表中的值插入PHP中的第三个空表

1 个答案: