登錄表單當電子郵件和密碼不匹配或電子郵件不存在時顯示消息

Question

我有一個登錄表單，它在密碼輸入錯誤但電子郵件地址不正確時顯示一條消息。但是當電子郵件或兩者都不正確時，它不會顯示任何內容。我希望它在缺少這兩種情況下顯示消息（「密碼和電子郵件不匹配」）。我真的很讚賞任何幫助。

<div id="container"> 
      <div> 
       <h1>Log In</h1> 
       <form method="POST" action="logIn.php"> 
        <label>Email</label> 
        <input name= "txtEmail"/> 
        <label>Password</label> 
        <input type="password" name="txtPass"/> 
        <input type= "submit" name= "submit" value="Log In"/> 
       </form> 
      </div> 
     </div> 

<?php 

    require ('connection.php'); 

      if (isset($_POST["txtEmail"])) 
      { 
      $email= $_POST["txtEmail"];//NO RECONOCE LOS TEXT FIELDS!! 
      } 
      else 
      {$email= ""; 
      echo "Email empty";} 

      if (!empty($_POST['txtPass'])) 
      { 
      $pass= $_POST['txtPass']; 
      } 
      else 
      {$pass= ""; 
      echo "Password empty.";} 

      // Enviar consulta 
       $instruction = "SELECT Password, Email FROM customer WHERE Email = '". $email . "'"; 
       $query = mysql_query ($instruction, $connection) 
       or die ("Fallo en la consulta");  
      // Mostrar resultados de la consulta 
       $nrows = mysql_num_rows ($query); 
       if ($nrows > 0) 
       { 
        $result = mysql_fetch_array ($query); 
        if (isset($result['Password'])) 
        { 
        $clave = $result['Password']; 
        //echo "Isset password works."; 
        } 
        else {$result= "";} 
        Email doesnt exist, or ;} 


        if ($clave == $pass){ 
         header('Location: welcomeLogIn.php'); 
        } 
        else{ 
         echo "Password or Email doesnt match."; 
        } 
       } 
    mysql_close($connection); 
    ?> 

Answer 1

感謝您提供您的代碼供我們審查。基於代碼，我可以看到你剛剛開始學習PHP。我已經用一種非常簡單的方法更新了你的代碼，來完成你所要求的。基本上，如果電子郵件或密碼爲空，那麼它將$ empty_value變量設置爲1.然後在腳本結尾檢查$ empty_value變量是否等於1.如果這是您希望的消息打印。請注意，您的代碼存在許多安全問題，建議您閱讀以下指南中的所有信息：http://phpsec.org/projects/guide/。

<?php 

    require ('connection.php'); 
$empty_value = 0; 

      if (isset($_POST["txtEmail"])) 
      { 
      $email= $_POST["txtEmail"];//NO RECONOCE LOS TEXT FIELDS!! 
      } 
      else 
      {$email= ""; 
      $empty_value = 1;} 

      if (!empty($_POST['txtPass'])) 
      { 
      $pass= $_POST['txtPass']; 
      } 
      else 
      {$pass= ""; 
      $empty_value = 1;} 

      // Enviar consulta 
       $instruction = "SELECT Password, Email FROM customer WHERE Email = '". $email . "'"; 
       $query = mysql_query ($instruction, $connection) 
       or die ("Fallo en la consulta");  
      // Mostrar resultados de la consulta 
       $nrows = mysql_num_rows ($query); 
       if ($nrows > 0) 
       { 
        $result = mysql_fetch_array ($query); 
        if (isset($result['Password'])) 
        { 
        $clave = $result['Password']; 
        //echo "Isset password works."; 
        } 
        else {$result= "";} 
        Email doesnt exist, or ;} 


        if ($clave == $pass){ 
         header('Location: welcomeLogIn.php'); 
        } 
        else{ 
         echo "Password or Email doesnt match."; 
        } 
       } 
    mysql_close($connection); 
    ?> 

<?php 
if ($empty_value == 1) { 
echo "Password or Email doesn't match."; 
} 
?> 

    <div id="container"> 
      <div> 
       <h1>Log In</h1> 
       <form method="POST" action="logIn.php"> 
        <label>Email</label> 
        <input name= "txtEmail"/> 
        <label>Password</label> 
        <input type="password" name="txtPass"/> 
        <input type= "submit" name= "submit" value="Log In"/> 
       </form> 
      </div> 
     </div> 

Answer 2

如果沒有匹配的電子郵件，那麼你會得到任何結果，所以你可以簡單的添加：

if ($nrows > 0) 
{ 
    //You current code to check if the password is correct     
} 
else{ 
    echo "No matching email found"; 
}