PHP调用函数失败

单击“提交”按钮时调用php函数时出现问题：

<?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
            }
        }
    }
    ?>

---

请帮帮我，我是php新手，这让我很恼火，从昨天晚上开始我就一直在想办法解决这个问题。我真的不知道有什么问题。我如何更改代码？除了friend_request（）函数部分之外，一切都很好。你的表单中的

用户名
输入缺失，基于允许进入块的
username
值
if（isset（$\u GET['username']）==true，空（$\u GET['username']）==false）{


谢谢，但这只是一个好友请求按钮，它不需要文本类型。我在阻止中允许的用户名用于其他目的。由于您的代码需要用户名，您可以将用户名作为隐藏值传递，如
仍然只需刷新页面添加到URL:Username？Username=Username&submit=Send+friend+request%21Am不是当然，我理解你的评论！
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')");
}}}

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