如何顯示消息，如果沒有結果返回數據庫查詢在PHP

Question

想知道是否有人可以幫助。我通過PHP和MySQL的教程運行，那裏是持有一個笑話數據庫：

• 笑話的笑話表
• 作者在作者表中類別表
• 分類

我有一個名爲authors.html.php的頁面，其中包含一個作者列表，其中包含添加作者，編輯/刪除作者（每個返回作者姓名旁邊）的選項。我所試圖做的是如果包含作者的數據庫是空的，而不是顯示下面的PHP錯誤：

Notice: Undefined variable: authors in C:\wamp\www\chapter7\admin\authors\authors.html.php on line 22

我想顯示一個漂亮整潔的消息說，沒有找到結果。

我的代碼如下：

<?php 
    include $_SERVER['DOCUMENT_ROOT'] . '/chapter7/includes/db.inc.php'; 
    try { 
      $result = $pdo->query('SELECT id, name FROM author'); 
    } catch (PDOException $e) { 
      $error = 'Error fetching authors from the database!'; 
      include $_SERVER['DOCUMENT_ROOT']. 
          '/chapter7/admin/error.html.php'; 
      exit(); 
    }  

    foreach ($result as $row){ 
      $authors[] = array('id' => $row['id'], 'name' => $row['name']); 
     } 

    include 'authors.html.php'; 

    if (isset($_POST['action']) and $_POST['action'] == 'Delete') { 
      include $_SERVER['DOCUMENT_ROOT']. 
          '/chapter7/includes/db.inc.php'; 
      //Get jokes belonging to author 
      try { 
       $sql = 'SELECT id FROM joke WHERE authorid = :id'; 
       $s = $pdo->prepare($sql); 
       $s->bindValue(':id', $_POST['id']); 
       $s->execute(); 
      } catch (PDOException $e) { 
       $error ='Error getting list of jokes to delete.'; 
       include $_SERVER['DOCUMENT_ROOT']. 
           '/chapter7/admin/error.html.php'; 
       exit(); 
      } 
      $result = $s->fetchAll(); 
      //Delete joke category entries 
      try { 
       $sql = 'DELETE FROM jokecategory WHERE jokeid =:id'; 
       $s = $pdo->prepare($sql); 
       //For each joke 
       foreach ($result as $row) { 
         $jokeId= $row['id']; 
         $s->bindValue(':id', $jokeId); 
         $s->execute();  
       } 
      } catch (PDOException $e) { 
       $error = 'Error deleting category entries for joke.'; 
       include $_SERVER['DOCUMENT_ROOT']. 
           '/chapter7/admin/error.html.php'; 
       exit(); 
      } 
      //Delete jokes belonging to author 
      try { 
       $sql = 'DELETE FROM joke WHERE authorid = :id'; 
       $s = $pdo->prepare($sql); 
       $s->bindValue(':id', $_POST['id']); 
       $s->execute(); 
      } catch (PDOException $e) { 
       $error = 'Error deleting jokes for author'; 
       include $_SERVER['DOCUMENT_ROOT']. 
           '/chapter7/admin/error.html.php'; 
       exit(); 
      } 
      //Delete the author 
      try { 
       $sql = 'DELETE FROM author WHERE id =:id'; 
       $s = $pdo->prepare($sql); 
       $s->bindValue(':id', $_POST['id']); 
       $s->execute(); 
      } catch (PDOException $e) { 
       $error ='Error deleting author'; 
       include $_SERVER['DOCUMENT_ROOT']. 
          '/chapter7/admin/error.html.php'; 
       exit(); 
      } 
      header('Location: .'); 
      exit(); 
    }  

請注意我非常新的這個和已經嘗試了一些反覆失敗的悲慘，所以任何幫助，甚至只是簡單的指導，將步驟我通過思考圍繞創建這樣的功能將不勝感激。不只是尋找正確的答案，而是想要了解如果任何人都可以提供幫助的背後的想法會很棒。

Answer 1

類似的東西應該工作。我過去曾經使用過它，它已經爲我工作。你將需要調整查詢等到你的數據庫和表格，但它應該沒問題。

$stm = $PdoObj->prepare("SELECT * FROM NEWS_articles"); 
    $stm ->execute(); 
    $results = $stm ->fetchAll(); 

    if($results==null){ 
     echo "<p>No dice, try other criteria</p>"; 
    }else{ 
     foreach($results as $row){ 
      echo $row["articleName"]; 
     } 
    }