所需输入字段的PHP错误消息

Question

如果用户没有输入电子邮件而没有向表中添加列，我正在尝试显示错误消息。

php代码发送电子邮件＆amp;随机生成的代码到数据库，然后显示一条消息，但如果他们没有输入电子邮件，我试图收到一条消息要求发送电子邮件，而不是创建一个列，直到完成。

我之前已经通过其他数据输入实现了这一点，但我还是PHP新手，我已经检查了答案，但不能找到我知道的简单修复。

非常感谢任何帮助或评论，谢谢（Y）

PHP：

 <?php
 if(!empty($_POST['email'])) {
 ?>

   <p>Enter an email</p>

 <?php
  }

 if(isset($_POST['send'])){

 $email = mysqli_real_escape_string($conn,$_POST['email']);

$length = 10;
$inviteCode = "";
$characters = "0123456789abcdefghijklmnopqrstuvwxyz";
for ($p = 0; $p < $length; $p++) {
$inviteCode .= $characters[mt_rand(10, strlen($characters))];
 }

$query = mysqli_query($conn, "INSERT INTO `referrals` (`email`, `inviteCode`) 
 VALUES ('$email', '$inviteCode') ");

if($query){
?>
  <p> "Thank you"</p>
<?php 

  }
  else{
?>

   <p>Sorry there must have been a problem</p>

<?php
    die('Error querying database. ' . mysqli_error($conn));
   }
  } // end brace for if(isset($_POST['Login']))
  // end brace for if(isset($_POST['Login']))
?>

我之前通过将以下内容放在首位来实现这一目标;

 if (empty($_POST) === false) {
    $required_fields = array('email');
     foreach($_POST as $key=>$value) {
       if (empty($value) && in_array($key, $required_fields) === true) {
        $errors[] = 'Please Enter All Information';
        break 1;
       }
    }
  }

提交和电子邮件输入标记的html代码如下：

    <input type="text" placeholder="Email" name="email" />  
  <input type="submit" value="send" name="send" />

Answer 1

将if(!empty($_POST['email']))添加到逻辑顶部，而不是isset($_POST['send'])，除非您在逻辑中稍后使用post send值执行其他操作。如果有一个空字符串，则isset将为true。

要了解原因，请参阅：

isset() and empty() - what to use

修改

<?php
if(!empty($_POST)){
    if(!empty($_POST['email'])){

        $email = mysqli_real_escape_string($conn,$_POST['email']);

        $length = 10;
        $inviteCode = "";
        $characters = "0123456789abcdefghijklmnopqrstuvwxyz";

        for ($p = 0; $p < $length; $p++) {
             $inviteCode .= $characters[mt_rand(10, strlen($characters))];
        }

         $query = mysqli_query($conn, "INSERT INTO `referrals` (`email`, `inviteCode`) VALUES ('$email', '$inviteCode') ");
         //you might want to consider checking more here such as $query == true as it can return other statuses that you may not want
         if($query){
         ?>
             <p> "Thank you"</p>
         <?php 
        }
        else{
        ?>
           <p>Sorry there must have been a problem</p>
        <?php
            die('Error querying database. ' . mysqli_error($conn));
        }
    }
    else {
    ?>
        <p>Enter an email</p>
    <?php
    }
}
?>

Answer 2

试试这个：

<?php
 // check if email is empty, if it is display this message
 if(empty($_POST['email'])) {
 ?>

   <p>Enter an email</p>

 <?php
 }
 // else: the user entered something
 else {

   $email = mysqli_real_escape_string($conn,$_POST['email']);

   $length = 10;
   $inviteCode = "";
   $characters = "0123456789abcdefghijklmnopqrstuvwxyz";
   for ($p = 0; $p < $length; $p++) {
     $inviteCode .= $characters[mt_rand(10, strlen($characters))];
   }

   $query = mysqli_query($conn, "INSERT INTO `referrals` (`email`, `inviteCode`) 
     VALUES ('$email', '$inviteCode') ");

   if($query){
?>
  <p> "Thank you"</p>
<?php 
   }
   else {
?>

   <p>Sorry there must have been a problem</p>

<?php
     die('Error querying database. ' . mysqli_error($conn));
   } // end else
 } // end else
?>