Question

一直试图从各种教程中学习如何创建用户输入表单并使用php发送到mysql。我最终从不同的来源学习代码的不同部分，并且经常在将它们组合在一起时发现这个错误。

我想我理解每个部分正在做什么，并且看不到任何语法错误，但毫无疑问它在某个地方是错误的......

请帮助它让我发疯（关于表格和安全mysqli的完整教程的任何建议，谢谢）

    <?php # DISPLAY COMPLETE REGISTRATION PAGE.

    # Set page title and display header section.
    $page_title = 'mysqli_form2' ;


    # Check form submitted.
    if ( $_SERVER[ 'REQUEST_METHOD' ] == 'POST' );
    {
      # Connect to the database.
    require ('../connect_db.php');

    // Escape user inputs for security
    $first_name = mysqli_real_escape_string($link, $_POST['first_name']);
    $last_name = mysqli_real_escape_string($link, $_POST['last_name']);
    $email_address = mysqli_real_escape_string($link, $_POST['email']);

    // attempt insert query execution
    $sql = "INSERT INTO mysql_test (first_name, last_name, email_address)VALUES('$first_name', '$last_name', '$email')";
    if(mysqli_query($link, $sql))
    {
    echo "success.";
     } 
    else
    {
    echo "ERROR: not able to execute $sql. " . mysqli_error($link);
     }

     // close connection 
     mysqli_close($link);



     ?>



     <!-- Display body section with sticky form. -->
     <h1>stu's test form...</h1>
     <form action="mysqli_form2.php" method="post">

     <p>First Name: <input type="text" name="first_name" size="20" 
     value="<?php if (isset($_POST['first_name'])) echo POST['first_name'];?"> 
     <p>Last Name: <input type="text" name="last_name" size="20" 
      value="<?php if (isset($_POST['last_name'])) echo$_POST['last_name'];?"></p>

      <p><input type="submit" value="GO...."></p>
      </form>

解析错误：语法错误，第44行的C：\ Abyss Web Server \ htdocs \ mysqli_form2.php中的意外文件结束

0 个答案: