PHP IF/ELSE條件不能按預期工作

Question

我試圖實現多個if條件來檢查提案的狀態，並且只應執行特定代碼（如果狀態代碼設置爲1或5）。

由於某種原因，我在實施這方面遇到了困難。目前代碼中的邏輯是，如果提案狀態不匹配1或5，則返回消息，否則執行下一個查詢。當我只指定一個數字，即（1或5）時，它會正常工作。

我與若和其他條件，在這部分面臨着另一個問題：

if ($count == 1) { 

     $feedback = '<p class="text-danger"> You have already accepted an application. You cannot accept or apply for any others. If this is a mistake then please contact the supervisor concerned directly.</p>'; 
    } 

    if ($count < 1) { 

     $status = $db_conx->prepare ("SELECT status_code FROM record WHERE student_record_id = :user_record_id AND proposal_id = :proposal"); 

     $status->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR); 
     $status->bindParam(':proposal', $proposal, PDO::PARAM_STR); 
     $status->execute(); 

     $proposalstatus = $status->fetchColumn(); 

     if($proposalstatus != 1) 
     { 
       //echo $proposalstatus; 
      $feedback = '<p class="text-danger">The proposal is not at a status where it can be accepted</p>'; 
     } 
    } 

    else { 

當我單獨運行每個部分，但是當我試圖把它一起在if語句中失敗，並沒有按這個工程根本不檢查這些條件，只是完成更新數據庫並顯示成功消息的任務。

完整的代碼是在這裏：

try 
    { 
    $db_conx = new PDO("mysql:host=$mysql_hostname;dbname=$mysql_dbname", $mysql_username, $mysql_password); 

    $db_conx->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION); 

    $username = $_SESSION['username']; 

    $sql = $db_conx->prepare("SELECT username, user_record_id FROM login_details 
     WHERE username = :username"); 

    $sql->bindParam(':username', $username, PDO::PARAM_STR); 

    $sql->execute(); 
    $user_record_id = $sql->fetchColumn(1); 

    $proposal = $_POST['proposal_id']; 

    $acceptCheck = $db_conx->prepare ("SELECT * FROM record WHERE student_record_id = :user_record_id AND status_code = 3"); 
    $acceptCheck->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR); 
    $acceptCheck->execute(); 

    $count = $acceptCheck->rowCount(); 

    if ($count == 1) { 

     $feedback = '<p class="text-danger"> You have already accepted an application. You cannot accept or apply for any others. If this is a mistake then please contact the supervisor concerned directly.</p>'; 
    } 

    if ($count < 1) { 

     $status = $db_conx->prepare ("SELECT status_code FROM record WHERE student_record_id = :user_record_id AND proposal_id = :proposal"); 

     $status->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR); 
     $status->bindParam(':proposal', $proposal, PDO::PARAM_STR); 
     $status->execute(); 

     $proposalstatus = $status->fetchColumn(); 

     if($proposalstatus != 1 || 5) //status must be either 'Approved' code 1 or 'Held' code 5 
     { 
       //echo $proposalstatus; 
      $feedback = '<p class="text-danger">The proposal is not at a status where it can be accepted</p>'; 
     } 
    } 

    else { 

       //Update all application records to 'Not available' when a proposal has been accepted 

     $updateOtherRecords = $db_conx->prepare("UPDATE record SET status_code = 8, last_updated = now() 
      WHERE proposal_id = :proposal"); 
     $updateOtherRecords->bindParam(':proposal', $proposal, PDO::PARAM_STR); 
     $updateOtherRecords->execute(); 

       //Update other applicationa for the user concerned to 'Rejected' 

     $updateUserRecord = $db_conx->prepare("UPDATE record SET status_code = 7, last_updated = now() 
      WHERE student_record_id = :user_record_id"); 
     $updateUserRecord->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR); 
     $updateUserRecord->execute(); 

       //Update the proposal concerned and assign it to the user 

     $update = $db_conx->prepare("UPDATE record SET status_code = 3, last_updated = now() 
      WHERE proposal_id = :proposal AND student_record_id = :user_record_id"); 
     $update->bindParam(':user_record_id', $user_record_id, PDO::PARAM_STR); 
     $update->bindParam(':proposal', $proposal, PDO::PARAM_STR); 
     $update->execute(); 

     $feedback = '<p class="text-success"> The proposal has been successfully accepted <span class="glyphicon glyphicon-ok"/></p>'; 
    } 
} 

我真的需要知道我怎麼可以排序，因爲我將使用if和else很多這種說法。任何指導將非常感謝！

預先感謝您！

Answer 1

你的條件不

if ($count < 1) { 
    some stuff 
} 

if ($count == 1) { 
... 
} else 
... this code will execute when $count is *NOT* equal to 1, 
    which includes when it's LESS than 1, e.g. "< 1" is true here 
} 

也許你想

if ($count == 1) { 
} else if ($count < 1) { 
} else { 
} 

讓別人最終如果/當$count >= 1

Answer 2

與提案狀態替換你的病情纔會運行互斥1或5

if(!($proposalstatus == 1 || $proposalstatus == 5)) { 
$feedback = '<p class="text-danger">The proposal is not at a status where it can be accepted</p>'; 
}