Question

我需要创建一个表单，它将检查输入是否为空。如果它是空的，那么应该有一个像“必填字段”这样的文本。有一条通知说，surName有一个未定义的索引。

这是我的PHP代码

<?php
  $name_error="";
  $sname_error="";
  $f_name="";
  $s_name="";

  if (isset($_POST['submit_button'])) {
    if ($_POST['firstName']!=='') {
      $f_name=$_POST['firstName'];
    }
    else {
      $name_error="Required Field *";
    }
  }
  if ($_POST['surName']!=='') {
    $s_name=$_POST['surName'];
  }
  else {
    $sname_error="Please fill this out";
  }
?>


 <html>
   <head>
     <title>Registration form</title>
   </head>
   <body>
     <div class="head">
       <p>Registration Form</p>
     </div>
     <form action="Register/final.php" method="POST">
       <label for="firstName">First Name</label><br>
       <input type="text" name="firstName" placeholder="First Name" value="<?php echo $f_name; ?>"><br><br>           
       <p style="color: red;"><?php echo $name_error; ?></p>
       <label for="surName">Last Name</label><br>
       <input type="text" name="surName" placeholder="Last Name" value="<?   
         php echo $s_name;?>"><br><br>
       <p style="color: red;"><?php echo $sname_error;?></p>
     </form>
   </body>
 </html>

Answer 1

对于初学者来说，你可以使用内置的HTML5输入验证器required

<input type="text" required>

这将在提交时提供好的用户消息，要求他们填写该字段。

其次，您可以使用empty代替if ($_POST['surName']!=='')

来检查服务器

if ( !empty($_POST['surName']) ) {
   // your logic here
}

Answer 2

功能基本上已经存在......

我想var possibleChoices = new Dictionary<string,string>{ {"1", "Latte"}, {"2", "Cappuccino"}, {"3", "Espresso"}, {"4", "Double espresso"} }; string value; var message = possibleChoices.TryGetValue(choice, out value) ? string.Format("You have selected: {0}", value) : "Incorrect value, please try again"; //In your code, the fall-through case misses the leading \n . //By avoiding repetition, consistency in formatting is achieved //with a single line of code Console.WriteLine("\n{0}", message);的{{1}}括号应该在}
缺少if (isset($_POST['submit_button'])) {
考虑 Jeff Puckett II 所说的

因此，如果我对代码进行了更改1和2，则可能如下所示：

?>

Answer 3

这是因为你把surName验证放在提交条件的外部; `

         if (isset($_POST['submit_button'])) {
              if ($_POST['firstName']!=='') {
                  $f_name=$_POST['firstName'];
                 }else{
                    $name_error="Required Field *";
                  }
                 if ($_POST['surName']!=='') {
                   $s_name=$_POST['surName'];
                  }
                else{
                  $sname_error="Please fill this out";
                 }
               }

     ?>`

您可以使用像errors = array();这样的关联数组，然后使用$errors['surName'] = "this is field is required"来改善这一点;

Answer 4

您没有提交按钮。另一件事是您提交后未使用if $_POST['surName']语句<?php error_reporting(1); $name_error = ""; $sname_error = ""; $f_name = ""; $s_name = ""; if (isset($_POST['submit_button'])) { if ($_POST['firstName'] != '') { $f_name = $_POST['firstName']; } else { $name_error = "Required Field *"; } if ($_POST['surName'] != '') { $s_name = $_POST['surName']; } else { $sname_error = "Please fill this out"; } } ?> <html> <head> <title>Registration form</title> </head> <body> <div class="head"> <p>Registration Form</p> </div> <form action="Register/final.php" method="POST"> <label for="firstName">First Name</label><br> <input type="text" name="firstName" placeholder="First Name" value="<?php echo $f_name; ?>"><br><br> <p style="color: red;"><?php echo $name_error; ?></p> <label for="surName">Last Name</label><br> <input type="text" name="surName" placeholder="Last Name" value="<?php echo $s_name;?>"><br><br> <p style="color: red;"><?php echo $sname_error;?></p> <input type="submit" name="submit_button" value="Register"> </form> </body> </html>。您可以像这样使用它：

TGA

表单验证问题PHP

4 个答案: