在PHP中显示错误消息的更好方法是什么？

问题描述：

我创建了三个php表单。第一个（form.php）包含表单标签，第二个（form1.php）是信息所在的位置。最后一个（formerror.php），我创建时，我找不到的方式在第一页（form.php）上显示一条错误消息。我想知道的是，这样做是否合适？ - 创建与第一个完全相同的页面以显示错误消息。在PHP中显示错误消息的更好方法是什么？

这是第一页。

这是最后一页，我必须准确地在第一页，以显示错误消息复制。

第一页有从我进入信息，并在第二页是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

一旦从form.php的，这里的到来，进入信息显示得到，因为我们给他们一种形式已经给出了用户名和电子邮件地址代码的第二页（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>

，这是我为了显示错误创建页面。我所做的是，我复制了与form.php完全相同的页面，但是在底部，我添加了一条消息，如图所示。

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，页面在那里。而且我也想知道我的代码是否做得很好，虽然他们完美地工作。

答

试试这个。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; 
}

答

您可以通过仅使用一个文件来解决问题。将您的表单数据发送到同一页面。像这样在你的文件开始之前添加你的PHP脚本。

<?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; 
    } 
?>

然后添加您的html代码。

<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>

在PHP中显示错误消息的更好方法是什么？

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