操作页面更改时，表单不会提交

Question

我有一个带有表单的简单页面(如下所示)。如果action='ask.php‘，表单将提交。(ask.php是当前页面，下面是代码)。当我将ask.php更改为question.php (文件夹中的另一个页面)时，表单不会提交。为什么会这样呢？是否在将数据放入数据库之前重定向表单？

ask.php代码；

<?php session_start(); ?>
       <!DOCTYPE html>
      <html xmlns="http://www.w3.org/1999/xhtml">
        <head>
        <title></title>

      <script type='text/javascript' src='http://code.jquery.com/jquery-1.7.1.min.js'></script>
      <script type='text/javascript' src='http://twitter.github.com/bootstrap/1.4.0/bootstrap-modal.js'></script>
        <link rel="stylesheet" href="../css/bootstrap.css" type="text/css" media="screen" />

         </head>
          <body>
       <?php include '../css/bar.php'; ?>
    <div id='content'>
<?php include '../nav.php'; 
?>



<?php


if (isset($_POST['submit'])){
include '../connect.php';
$question=mysql_real_escape_string($_POST['question']);
$detail=mysql_real_escape_string($_POST['detail']);
$date=date("d M Y");
$time=time();
$user=$_SESSION['id'];
$put=mysql_query("INSERT INTO questions VALUES ('','$question','$detail','$date','$time','$user','subject','0')");

$result=mysql_query("SELECT * FROM questions WHERE user='$user' AND time='$time'");
    while ($row = mysql_fetch_assoc($result)){

         $q=$row['id'];

        }
}

?>

<form method='POST' action='question.php'>   //won't submit because action is question.php?
    <p>Question:</p>
    <p><input type='text' name='question' id='question'  maxlength='200'></p>
    <p>Add some detail (optional):</p>
    <p><textarea id='detail' name='detail' ></textarea></p>

    <p><input type='submit' value='submit' name='submit'></p>
</form>


</div>
<?php include '../footer.php'; ?>
</body>

​

Answer 1

Jeroen是对的：

操作页面应包含此类信息，以便将信息发送到数据库：

第一部分和第二部分准备将数据插入到数据库中：

$question = mysql_real_escape_string($_POST['question']); // Part 1
$detail = mysql_real_escape_string($_POST['detail']); // Part 2
$con = mysql_connect("server","username","password");
mysql_select_db("YourDatabase", $con);

$sql="INSERT INTO YourTable (question, details) VALUES ('$question','$detail')";
if (!mysql_query($sql,$con)) {
    //Some error handling here,
    // Stay away from using die in all but development
}
mysql_close($con)

您也应该使用preg_match测试格式，以检查数据是否为预期的格式。希望这能有所帮助

编辑！

这种连接方式已经过时了！使用PDO进行数据库连接。一些文档

PDO Article

PDO php docs