PHP：程序无法运行。我不明白错误

我已经创建了这3页的表格。第一是注册，第二是入境和出境日期，现在是第三次，我想让用户在价值停车，迎接和迎接服务停车和灵活的停车场之间进行选择。我在他们每个人旁边创建了一个“选择”按钮，但是，当我点击按钮时，我检查了我的数据库，并且只有id和用户名得到更新，停车名不会出现..这里是我的代码：PHP：程序无法运行。我不明白错误

<?php 
session_start(); 
$error=true; 
$users=$_SESSION['tsmUserName']; 
$Parkingname=""; 
if (isset($_POST['tsmValueaParking'])){ 
    $name="value Parking"; 
    $error=false; 
} 
if (isset($_POST['tsmMeetGreetService'])){ 
    $name="Meet and Greet Service Parking"; 
$error=false; 
} 
if (isset($_POST['tsmFlexible'])){ 
    $name="Flexible Parking"; 
    $error=false; 
} 
if($error==false){ 
    require_once("connection.php"); 
      $my_query="INSERT INTO `parking`(`Id`, `name`, `username`) VALUES (NULL,'$name','$users')"; 
      $result=mysqli_query($connection,$my_query); 
      if($result) 
      { 
       echo 'thank you'; 
      } 
      else 
      { 
       echo 'error'; 
      } 
      mysqli_close($connection); 
} 

?> 
<html> 
<head></head> 
<body> 
<form name="picking" id="picking" method="POST" action=""> 
<table cellpadding="5" border="0" width="100%"> 
<tr> 
      <td align="center"> 
       <h2> Welcome to Tassimo Online Supermarket </h2> 
       <hr> 
      </td> 
     </tr> 
     <tr> 
     <td width="30%" align="left"> 
        <label for="tsmdate">Value Parking</label></br></br> 
        Our car park is the closest off car park to both terminals. FREE shuttle bus runs every 10 minutes and gets you to the terminals in 3-5 minutes. 
       </td> 
       </tr> 
       <tr> 
       <td align="left"> 
        <input type="button" name="tsmValueaParking" id="tsmValueaParking" value="pick"> 
       </td> 

       </tr> 

       <tr> 
       <td width="30%" align="left"> 
        <label for="tsmdate">Meet & Greet Service</label></br></br> 
        This is a Meet and Greet Service where you will be met at your departure terminal. Your car will be parked at the secure facility. With this service staff member will meet you at your departure terminal and take your car back to their secure compound. On your return staff member will meet you at the Terminal with your car so you can be on your way home as quickly as possible. 
       </td> 
       </tr> 
      <tr> 
       <td align="left"> 
        <input type="button" name="tsmMeetGreetService" id="tsmMeetGreetService" value="pick"> 
       </td> 
       </tr> 

       <tr> 
       <td width="30%" align="left"> 
        <label for="tsmdate">Flexible</label></br></br> 
        This car park is located just 500 meters away. The bus service runs every 30 mins. Tickets cost $4.50 one way and are available from the driver. 
       </td> 
       </tr> 
       <tr> 
       <td align="left"> 
        <input type="button" name="tsmFlexible" id="tsmFlexible" value="pick"> 
       </td> 

       </tr> 
      </table> 


</form>

来源

2016-12-17 Hamza Khan

在你的''Parkingname =“”;'在最上面的行中加入'$ name ='test';'看看它是否添加到数据库中。此外，请执行'print_r（$ _ POST）;'看看您提交的内容是否符合您的期望。 – Rasclatt

@Rasclatt我试过var_dump，但它无法识别$ _POST，因为数组显示为0 ..我该怎么做：/我很困惑 –

将所有' Rasclatt

该问题可能是由于您的表单由于字段类型而未正确提交。将所有您：

<input type="button"

到

<input type="submit"

然后尝试并再次提交。我认为如果切换类型提交您将看到您的$_POST正常工作。

在附注中，如果您有用户提交的数据，则需要确保仔细绑定参数。你想避免sql注入。

来源

2016-12-17 06:29:28 Rasclatt

PHP：程序无法运行。我不明白错误

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