在PHP中顯示錯誤消息的更好方法是什麼？

我創建了三個php表單。第一個（form.php）包含表單標籤，第二個（form1.php）是信息所在的位置。最後一個（formerror.php），我創建時，我找不到的方式在第一頁（form.php）上顯示一條錯誤消息。我想知道的是，這樣做是否合適？ - 創建與第一個完全相同的頁面以顯示錯誤消息。在PHP中顯示錯誤消息的更好方法是什麼？

這是第一頁。

這是最後一頁，我必須準確地在第一頁，以顯示錯誤消息複製。

第一頁有從我進入信息，並在第二頁是form1.php

CODE FROM THE FIRST PAGE(form.php): 
<!DOCTYPE html> 
<html> 
<head> 
    <title>Form</title> 
    <style type="text/css"> 
     input { 
      display: block; 
      margin: 10px; 
      padding: 10px; 
     } 
     form { 
      border: 0px solid rgba(0,0,0,0.5); 
      width: 19%; 
      padding: 20px; 
      background-color: rgba(5,25,9,0.2); 
     } 
    </style> 
</head> 
<body> 
//This is the form from form.php 
    <form action="form1.php" method="post"> 
     <input type="text" name="username" placeholder="Enter username"> 
     <input type="email" name="email" placeholder="Enter email address"> 
     <input type="password" name="password" placeholder="Enter password"> 
     <input type="password" name="cpassword" placeholder="Enter password again"> 
     <input type="submit" name="submit_btn" value="Submit"> 
     <input type="reset"> 
    </form> 

    <p style="color: red;" class="erreur"></p> 


    </body> 
    </html> 

    //This is the second page

一旦從form.php的，這裏的到來，進入信息顯示得到，因爲我們給他們一種形式已經給出了用戶名和電子郵件地址代碼的第二頁（form1.php）：

<!DOCTYPE html> 
    <html> 
    <head> 
     <title>FORM 1</title> 
    </head> 
    <body> 

    <?php 
     $username = $_POST['username']; 
     $email = $_POST['email']; 
     $submit_btn = $_POST['submit_btn']; 
     $password = $_POST['password']; 
     $cpassword = $_POST['cpassword']; 

     if (isset($_POST['submit_btn'])) { 
      if ((empty($_POST['username'])) || (empty($_POST['email'])) || (empty($_POST['password'])) || (empty($_POST['cpassword']))) { 
       // header("Location:form.php"); 
       header("Location:formerror.php"); 
      }else{ 
       echo "Welcome $username, your email address is $email"; 
      } 
     } 

    ?> 



    </body> 
    </html>

，這是我爲了顯示錯誤創建頁面。我所做的是，我複製了與form.php完全相同的頁面，但是在底部，我添加了一條消息，如圖所示。

AND CODE FROM THE THIRD PAGE(formerror.php): 

    <!DOCTYPE html> 
    <html> 
    <head> 
     <title>Form</title> 
     <style type="text/css"> 
      input { 
       display: block; 
       margin: 10px; 
       padding: 10px; 
      } 
      form { 
       border: 0px solid rgba(0,0,0,0.5); 
       width: 19%; 
       padding: 20px; 
       background-color: rgba(5,25,9,0.2); 
      } 
     </style> 
    </head> 
    <body> 

    <form action="form1.php" method="post"> 
     <input type="text" name="username" placeholder="Enter username"> 
     <input type="email" name="email" placeholder="Enter email address"> 
     <input type="password" name="password" placeholder="Enter password"> 
     <input type="password" name="cpassword" placeholder="Enter password again"> 
     <input type="submit" name="submit_btn" value="Submit"> 
     <input type="reset"> 
    </form> 

    <p style="color: red;">Please fill the form</p> 


    </body> 
    </html>

正如你所看到的，頁面在那裏。而且我也想知道我的代碼是否做得很好，雖然他們完美地工作。

來源

2017-04-25 Ayno

試試這個。放在你的php文件上面。

$server = "localhost"; 
$username= "root"; 
$pass = ""; 
$dbname =""; 
$conn = new mysqli($server, $username, $pass, $dbname); 

if($conn -> connect_error){ 
    echo "Connection Failed!, " . $conn -> connect_error; 
}

來源

2017-04-25 09:58:25

您可以通過僅使用一個文件來解決問題。將您的表單數據發送到同一頁面。像這樣在你的文件開始之前添加你的PHP腳本。

<?php   
    $error = ''; 
    if (isset($_POST['submit_btn'])) 
    { 
     $server = "localhost"; 
     $username= ""; // your username 
     $pass = ""; // your password 
     $dbname =""; // your database name 
     $conn = new mysqli($server, $username, $pass, $dbname); 
     if($conn -> connect_error){ 
      echo "Connection Failed!, " . $conn -> connect_error; 

     if(checkValidation()) 
      // insert the data into the database and redirect to a new page.      
    } 

    function checkValidation() 
    { 
     $username = $_POST['username']; 
     $email = $_POST['email']; 
     $password = $_POST['password']; 
     $cpassword = $_POST['cpassword']; 
     if(empty($username)) 
     { 
      $error = "Username field can't be empty";  
      return false;      
     } 
     if(empty($email)) 
     { 
      $error = "Email field can't be empty";  
      return false;      
     } 
     if(empty($password)) 
     { 
      $error = "Password can't be empty";       
      return false; 
     } 
     if(empty($cpassword) || ($password != $cpassword)) 
     { 
      $error = "Passwords don't match";       
      return false; 
     } 
      return true; 
    } 
?>

然後添加您的html代碼。

<html> 
<head> 
    <title>Form</title> 
    <style type="text/css"> 
     input { 
      display: block; 
      margin: 10px; 
      padding: 10px; 
     } 
     form { 
      border: 0px solid rgba(0,0,0,0.5); 
      width: 19%; 
      padding: 20px; 
      background-color: rgba(5,25,9,0.2); 
     } 
    </style> 
</head> 
<body> 
<?php> 
    if(!empty($error)) 
    { 
     echo "<p style='color: red;'></p>"; 
    } 
?> 
<form action="" method="post"> 
    <input type="text" name="username" placeholder="Enter username"> 
    <input type="email" name="email" placeholder="Enter email address"> 
    <input type="password" name="password" placeholder="Enter password"> 
    <input type="password" name="cpassword" placeholder="Enter password again"> 
    <input type="submit" name="submit_btn" value="Submit"> 
    <input type="reset"> 
</form> 
</body> 
</html>

來源

2017-04-25 10:27:46

在PHP中顯示錯誤消息的更好方法是什麼？

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