PHP失败，调用函数

Question

我有一个问题调用PHP函数，当我点击提交按钮：

<?php 
if (isset($_GET['username']) === true and empty($_GET['username']) === false) { 
$username  = $_GET['username']; 

if(user_exists($username) === true) { 
$profile_user_id  = user_id_from_username($username, 'username'); 
$my_id     = $_SESSION['user_id']; 
$profile_data   = user_data($profile_user_id, 'username', 'first_name', 'last_name', 'email', 'profile'); 
?> 

    <h1><?php echo $profile_data['first_name']; ?>'s Profile</h1> 
    <?php 
    if($profile_user_id != $my_id) { 
     $check_friend_query = mysql_query("SELECT id FROM friends WHERE (user_one='$my_id' AND user_two='$profile_user_id') OR (user_one='$profile_user_id' AND user_two='$my_id')"); 
     if(mysql_num_rows($check_friend_query) > 0) { 
      echo '<a href="">Already Friends </a>- <a href="">Unfriend '. $profile_data['username'] .'</a>'; 
     } else { 
      $from_query = mysql_query("SELECT id FROM friend_request WHERE `from` = '$profile_user_id' AND `to` = '$my_id'"); 
      $to_query = mysql_query("SELECT id FROM friend_request WHERE `from` = '$my_id' AND `to` = '$profile_user_id'"); 
      if(mysql_num_rows($from_query) == 1) { 
       echo '<a href="#">Ignore</a> or <a href="">Accept</a>'; 
      } elseif(mysql_num_rows($to_query) == 1){ 
       echo '<a href="#">Cancel Request</a>'; 
      } else { 
       if(isset($_GET['submit'])) { 
        friend_request(); 
        header('Location: '.$profile_user_id); 
        exit(); 
       } 
        ?> 
         <form action=""> 
          <input type="submit" name="submit" value="Send friend request!"> 
         </form> 
        <?php 
      } 
     } 
    } 
    ?> 

这最后还有ISSER（$ GET [“提交”]和函数调用form.What是错误我不能找到解决办法，它只是刷新页面，但doenst发送的mysql_query功能，我称之为是这样的：

function friend_request() { 
if (isset($_GET['username']) === true and empty($_GET['username']) === false) { 
$username  = $_GET['username']; 

    if(user_exists($username) === true) { 
    $profile_user_id  = user_id_from_username($username, 'username'); 
    $my_id     = $_SESSION['user_id']; 
    $profile_data   = user_data($profile_user_id, 'username', 'first_name', 'last_name', 'email', 'profile'); 

mysql_query("INSERT INTO friend_request VALUES('', '$my_id', '$profile_user_id')"); 
}}} 

请帮助我，即时通讯新的PHP，这惹恼了我，我一直停留在这个因为昨天晚上试图弄清楚。我真的没有看到问题是什么。如何我更改我的代码吗？一切都完美地期望friend_request（）函数的一部分。

Answer 1

在表单 username输入缺少的基础上， username值

已获准进入该块if (isset($_GET['username']) === true and empty($_GET['username']) === false) {

<form action=""> 
     <input type="text" name="username" value="your value"> 
     <input type="submit" name="submit" value="Send friend request!"> 
    </form>