PHP仅在第二次单击后才能工作

Question

我有这个PHP代码的代码：

$friendbutton;  
$sqla = "SELECT * FROM friends WHERE user1='$usernotif' AND user2='$username' AND accepted='0'";
$query = mysqli_query($link,$sqla) or die(mysqli_error($link)); 
$row_counter = mysqli_num_rows($query);
if ($row_counter > 0) {
    $friendbutton ='<input id="Friend" type="submit" name="Delete" value="Unsend friend request"><br />';
    $type="Unsend";
    $_SESSION['type']=$type;
} else {
    $friendbutton='<input id="Friend" name="Send" type="submit" value="Send friend request"><br />';
}

而这在我的friend.php ：

if (isset($_SESSION['Send'])){
    $sql = "INSERT INTO friends(user1, user2, datemade) VALUES('$usernotif','$username',now())";
    $sqla = "INSERT INTO notifications(username,initiator,date_time,type) VALUES('$username','$usernotif',now(),'Friend')";
}

但是，第二次单击后，“发送”和“删除”均起作用。

Answer 1

尝试这个：

<?php
$sqla = "SELECT * FROM friends WHERE user1='$usernotif' AND  user2='$username'
AND accepted='0'";

$query = mysqli_query($link, $sqla) or die(mysqli_error($link));
$row_counter = mysqli_num_rows($query);

if ( !empty($row_counter) ){
    $sql = "INSERT INTO friends(user1, user2, datemade) VALUES('$usernotif','$username',now())";
    $sqla = "INSERT INTO notifications(username,initiator,date_time,type) VALUES('$username','$usernotif',now(),'Friend')";
}
elseif (isset($_FORM['delete'])){
    //delete query goes here
}

?>

<form action="" method="post" accept-charset="utf-8">
<?php
    if ( !empty($row_counter) )
        echo '<input type="submit" name="Delete" value="Unsend friend request"><br />';
    else
        echo '<input type="submit" name="Send" value="Send friend request"><br />';
?>
</form>