未捕獲的異常錯誤PHP

致命錯誤：未捕獲的異常「異常」有消息「未捕獲的異常錯誤PHP

錯誤：以下字段尚未填寫OUT-姓」在/vagrant/web/Assignment4/Person.php上線9

我想檢查一個表單，以確保所有的字段都被填充。如果它們中的任何一個都是空的，我想拋出一個錯誤，指出哪些字段是空的。我很難理解如何捕捉異常，所以任何人都可以告訴我他們如何解決這個問題？

Person.php

public function insert() 
{ 

    //Storing required $_POST array fields in variable for isset function  

    $expectedValues = array(
     "firstName" => "First Name", 
     "lastName" => "Last Name", 
     "title"  => "Title", 
     "office"  => "Office", 
     "phoneNumber" => "Phone Number", 
     "email"  => "Email", 
     ); 


    //Checking to see if all fields are filled in 

    foreach ($expectedValues as $field => $humanName) { 
     if (empty($_POST[$field])) { 
     $errorArray[] = $humanName; 
     foreach($errorArray as $print){ 
      throw new Exception("<p>" . "Error: The following fields have not been filled out- " . $print . "</p>"); 
     } 

     try{ 
      (count($errorArray) = 0); 
     } 
     catch(Exception $e){ 
      echo "<p>" . $e->getMessage() . "</p>"; 
     } 
     } 
    } 


     //If they are, insert them into the Perosn table  

       $insert = $this-> doQuery("INSERT INTO Person 
             VALUES(
             '$_POST[firstName]', 
             '$_POST[lastName]', 
             '$_POST[title]', 
             '$_POST[office]', 
             '$_POST[phoneNumber]', 
             '$_POST[email]')"); 

            $insert; 


     //If insert query is successful, return true 
      if ($insert === true){ 
       return true; 
       echo "Congragulations! You now work for Assignment 4 Incorporated"; 
      } 


     //If not, throw an exception  

      //else{ 
       // throw new Exception 
       // ("<p>" . "Error: Query was unsuccessful:" 
       // . " " . $this->error . "</p>"); 
       // } 





    }

來源

2012-10-27 Brian Houlihan

如果你想顯示錯誤，然後用戶拋出異常是錯誤的方式，試試這個：

foreach ($expectedValues as $field => $humanName) { 
    if (empty($_POST[$field])) { 
     $errorArray[] = $humanName; 
    } 
} 
if (count($errorArray) > 0) { 
     echo 'Following fields are empty: '.implode(' ', $errorArray); 
}

而且爲了好玩，檢查出必要的HTML5特性：

<input type="text" required="required" value="" />

來源

2012-10-27 18:15:41

雖然不是（count（$ errorArray）> 0）嗎？是不是= 0，說如果錯誤數組中沒有任何內容可以打印出來？ –

謝謝，修復它 –

未捕獲的異常錯誤PHP

回答

相關問題