按钮取代另一个按钮取决于PHP查询

Question

我需要一个按钮来取代另一个按钮取决于几个PHP if statements。它的工作方式是，一个用户只能使用另一个用户一次，如果用户已经被使用，它将显示另一个按钮，而不是默认的一组按钮，这些按钮可用于非受欢迎的用户。

我在我的数据库中有一个表favourites。我插入了一行进行测试。

id - 16 
favourited_who - Freddy 
favourited_by - AliceP 

仅供参考，我的网站目前有三种用户：AliceP，Freddy和Fred。只是为了解释，我登录为AliceP，而且如您所见，Fred不在我的favourites中，在这种情况下，应该为Fred显示两个按钮，分别是Send Message和Add to Favourites。另一方面，Freddy是我的最爱，在这种情况下，我想显示两个按钮，它们是Send Message和Remove from Favourites。

使用下面的代码，Fred和Freddy将显示Remove from Favourites按钮，但不应该显示该按钮。

我需要覆盖在我的代码，我相信我有三个条件：

// 1. Con 1: Dont display any buttons for the logged in user as they cannot fav or msg themselves. 
// 2. Con 2: Display sendmsg and addfriend btn if the user is not in the logged in users favs 
// 3. Con 3: Display sendmsg and remfriend btn is the user is in the logged in users favs. 

这是我曾尝试：

if (isset($_POST['addfriend'])) { 
    $fav_request = $_POST['addfriend']; 
    $favourited_who = $user; // u variable 
    $favourited_by = $username; // logged in user 

    // Get all the favourites of the logged in user... 
    $q = mysqli_query ($connect, "SELECT * FROM favourites WHERE favourited_by='$username'"); 
    $num_of_favs = mysqli_num_rows($q); 
     $r_query = mysqli_fetch_array($q); 
      $db_fav_who = $r_query['favourited_who']; 
      $db_fav_by = $r_query['favourited_by']; 

    // Check: See if $user isn't already in there favourites. 
    /*if ($db_fav_by == $username){ // if logged in user has anyone favourited 
     // if the user already exists in the logged in users favourites, then display remove from favourites button. 
     if ($db_fav_who == $user){ 
      echo "<div class='edit_profile'> 
       <input type='submit' class='btn btn-info' name='remfriend' value='Remove from Favourites'> 
      </div>"; 
      } 
     }*/ 
     //}// while loop closed 
    if ($user != $username) { // Check: See user isnt favouriting themself. 
     $favourite_user = mysqli_query($connect, "INSERT INTO favourites VALUES ('', '$favourited_who', '$favourited_by')"); 
    } 
} // if isset closed 

// Condition 1: 
if ($user == $username){ 
     // dont display buttons  
} 
// Condition 2: 
if ($user != $username && $favourited_who != $db_fav_who){ // if $user is not equal to db fav_who value 
    echo "<form method='post'> 
       <input type='submit' class='btn btn-info' name='sendmsg' value='Send Message'/> 
       <input type='submit' class='btn btn-info' name='addfriend' value='Add to Favourites'> 
       </form>"; 
    }  
// Condition 3: 
if ($user != $username && $favourited_who == $db_fav_who) { 
     echo "<form method='post'> 
       <input type='submit' class='btn btn-info' name='sendmsg' value='Send Message'/> 
       <input type='submit' class='btn btn-info' name='remfriend' value='Remove from Favourites'> 
       </form>"; 
     } 

我已经覆盖了所有的三个条件，但我无法理解为什么代码没有为每个场景显示正确的按钮？

编辑：

找到了解决办法，原来我不得不添加if isset封闭括号外的以下内容：

$favourited_who = $user; // u variable 
    $favourited_by = $username; // logged in user 
    // Get all the favourites of the logged in user... 
    $q = mysqli_query ($connect, "SELECT * FROM favourites WHERE favourited_by='$username'"); 
    $num_of_favs = mysqli_num_rows($q); 
     $r_query = mysqli_fetch_array($q); 
      $db_fav_who = $r_query['favourited_who']; 
      $db_fav_by = $r_query['favourited_by']; 

Answer 1

你需要一个

<form method='post'> 

为以下行？

  echo "<div class='edit_profile'> 
      <input type='submit' class='btn btn-info' name='remfriend' value='Remove from Favourites'> 
     </div>"; 

其他有他们。

Answer 2

你写while循环关闭但循环在哪里？

可能

while($r_query = mysqli_fetch_array($q)) { ... }; 

而且你对此有何评论一个{和两个}在此代码

/* if ($db_fav_who == $user){ 
     echo "<div class='edit_profile'> 
      <input type='submit' class='btn btn-info' name='remfriend' value='Remove from Favourites'> 
     </div>"; 
     } 
    }*/