Question

我被要求创建在线注册表。我差点儿完成但是出现了一些问题。要注册活动，我创建了一个新页面，他们可以在其中查看活动的详细信息。但是，当文本写入文件大约一半时，它开始将输入写为文本而不是html / php代码。任何帮助将不胜感激。

 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Untitled Document</title>
</head>
<body>

<?php

   $sitename = $_POST[eName];
   $title = $_POST[eName];
   $sDate = $_POST[sDate];
   $enDate = $_POST[enDate];
   $exDate = $_POST[exDate];
   $desc = $_POST[desc];
   $pName = $_POST[eName];

   $con=mysqli_connect("localhost","emuas","flgdls","EMUAS_signUp");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$sql="INSERT INTO TEST (EventName, StartDate, EndDate, ExpiryDate, PageName, Description)
VALUES
('$_POST[eName]','$_POST[sDate]','$_POST[enDate]','$_POST[exDate]','$_POST[eName]','$_POST[desc]')";

if (!mysqli_query($con,$sql))
  {
  die('Error: ' . mysqli_error($con));
  }

  $eventID = mysqli_insert_id($con);

    $fileStart='<html>
   <head>
   <title>';

   $fileStTable=' </title>
   </head>
   <body>

 <table>
 <tr align="left">
 <th> ';

 $fileSDate=' </th>
 </tr>
 <tr align="left">
 <th> Start Date: </br>';

 $fileEnDate='</th>
 <th> End Date: </br>';

   $fileExDate='</th> 
 <th> Expiry Date: </br>';

 $fileDesc='</th>
 </tr>
 </table>
 <table id =createSheet>
 <tr align="left">
 <th> Description: </br>';

 $fileEnTable='</th>
 </tr>
</table>
';

 $fileStPHP='
 <?php 
$con=mysqli_connect("localhost","emuas","flgdls","EMUAS_signUp");
// Check connection
if (mysqli_connect_errno())
  {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
  }

$sign = mysqli_query($con,"SELECT * FROM SIGN_UP_TEST ORDER BY User");

echo "<table> 
<tr> 
<th>Name</th> 
<th>Status</th> 
<th>Comments</th> 
</tr>"; //Error Starts here. The "; are written as text and so is everything afterwards, until $fileStForm

while($row = mysqli_fetch_array($sign))
  {
  while($row["EventID"]=';


  $fileEnPHP='echo "<tr>";
  echo "<td>" . $row["User"] . "</td>";
  echo "<td>" . $row["Comments"] . "</td>";
  echo "</tr>";
  }
echo "</table>";

mysqli_close($con);
?>';

$fileStForm='
<form action="sign_user.php" method="post">
<input type="hidden" name="id" value="';

$fileEnForm='" />
<table>
<tr>
<td> Name: <br /> <input type="text" name="user" /> </td>
</tr>
<tr>
<td> Comments: <br /> <input type="text" name="comments" /> <input type="submit" value="Sign Up"/>  </td>
</tr>
</table>
</form>';

$fileEnd=' </body>
   </html>';


   $fp = fopen("sign_up_sheets/$sitename.html", "w");
   fputs($fp, $fileStart);
   fputs($fp, $title);
   fputs($fp, $fileStTable);
   fputs($fp, $sitename);
   fputs($fp, $fileSDate);
   fputs($fp, $sDate);
   fputs($fp, $fileEnDate);
   fputs($fp, $enDate);
   fputs($fp, $fileExDate);
   fputs($fp, $exDate);
   fputs($fp, $fileDesc);
   fputs($fp, $desc);
   fputs($fp, $fileEnTable);
   fputs($fp, $fileStPHP);
   fputs($fp, $eventID);
   fputs($fp, $fileEnPHP);
   fputs($fp, $fileStForm);
   fputs($fp, $eventID);
   fputs($fp, $fileEnForm);
   fputs($fp, $fileEnd);
   fclose($fp);

mysqli_close($con);

echo ("<SCRIPT LANGUAGE='JavaScript'>
    window.alert('Your sheet has been created')
    window.location.href='sign_up.php';
    </SCRIPT>");
    ?>


</body>
</html>

Answer 1

当然，正如戴文所写，你的MySQL并不正确（SQL注入）。我发现了一些其他问题。

您正在创建php脚本，但将其保存为html文件$fp = fopen("sign_up_sheets/$sitename.html", "w");（为什么？）您应该使用php，因为您的服务器没有将html扩展名解析为php脚本。这是正确的$fp = fopen("sign_up_sheets/$sitename.php", "w");（在我看来）
你的while循环中有一些错误。请查看您的$fileStPHP变量。你有while循环但只有外循环是正确的。也许你想要在那里声明？ while($row["EventID"]='; //inner loop ? But this hasn't got any logical explanation在我看来，这段代码应该与此类似：

while ($row = mysqli_fetch_array($sign)){
if($row["EventID"]==';
    $fileEnPHP=') {
        echo "<tr>";
        echo "<td>" . $row["User"] . "</td>";
        echo "<td>" . $row["Comments"] . "</td>";
        echo "</tr>";
    }
}

在php中写一个新的静态html页面

1 个答案: