表单页面上的PHP表单错误消息

Question

我对PHP还是很陌生，但似乎找不到我想要的东西。 我希望错误显示在填写表单的页面上。 我希望在表单页面的$ showerror部分显示“请输入名称”错误。

这是我到目前为止所拥有的...

<form method="post" action="process.php">
<table width="667">
<td colspan="2"><?php echo $showerror; ?></td>
<tr>
<td>Full Name:*</td>
<td><input class="text" name="name" placeholder="Ex. John Smith"></td>
<tr>
<td>E-mail:*</td>
<td><input class="text" name="email" placeholder="Ex. johnsmith@gmail.com"></td>
<tr>
<td>Type of site:*</td>
<td>Business<input name="multioption[]" type="radio" value="Business" class="check" /> Portfolio<input name="multioption[]" type="radio" value="Portfolio" class="check" /> Forum<input name="multioption[]" type="radio" value="Forum" class="check" /> Blog<input name="multioption[]" type="radio" value="Blog" class="check" /></td>
<tr>
<td>Anti-Spam: (2)+2=?*</td>
<td><input name="human" placeholder="What is it?"></td>
</table>
<input id="submit" name="submit" type="submit" value="Submit"></form>

这就是我的process.php的样子...我不确定如何编码错误部分。

<?php
$name         = isset($_POST['name'])         ? $_POST['name']         : '';
$email        = isset($_POST['email'])        ? $_POST['email']        : '';
$human        = isset($_POST['human'])        ? $_POST['human']        : '';
$submit       = isset($_POST['submit'])       ? true                   : false;

$multioption = isset($_POST['multioption'])
         ? implode(', ', $_POST['multioption'])
         : 'No multioption option selected.';



$from = 'From: Testing Form'; 
$to = 'xx@xxxxx.com'; 
$subject = 'Testing Form';

$body = 
"Name: $name\n 
E-Mail: $email\n 
Multi Options: $multioption\n";

if ($submit && $human == '4') {
    mail ($to, $subject, $body, $from);
    print ("Thank you. We have received your inquiry.");

}
else {
   echo "We have detected you are a robot!.";
    }
?>

Answer 1

您需要在PHP标记之间放置PHP语法，例如，以下行：

<td colspan="2">$showerror</td>

变成这样：

<td colspan="2"><?php echo $showerror; ?></td>

如果您完全不熟悉PHP，则可以从一些教程网站开始学习，这是一个不错的选择

编辑：

您可以在PHP页面中设置$ showerror，这可能是每个表单字段的一长串“如果条件”，但是我将向您展示一个小/简单的全名$_POST['name']示例，它将是这样的：

$showerror = '';
if(!empty($_POST['name'])) {// enter here if fullname is set
    if(strlen($_POST['name']) >= 6 && strlen($_POST['name']) <= 12) {// enter here if fullname length is between 6-12 characters
        // You can do more validation by using "if conditions" here if you would like, or keep it empty if you think fullname is correct
    } else {// enter here if fullname is NOT between 6-12 characters
        $showerror = 'Full name must be 6-12 characters';
    }
} else {// enter here if fullname is not set
    $showerror = 'Please enter your full name';
}

Answer 2

编译器仅将内容解析为<?php ?>标记之间编写的php，因此使用

<td colspan="2"><?php echo $showerror; ?></td>

要么

<td colspan="2"><?= $showerror ?></td>

Answer 3

尝试这个，

 <?php
    if(isset($showerror))
       echo $showerror;
    else 
       echo '';
  ?>

如果您没有在$showerror的第一次编译中验证$showerror变量的代码，则会收到错误消息Undefined variable

Answer 4

试试这个..您必须将两个代码都放在同一个文件中。 并可以检查请求是否发布仅读取php。 然后可以将值放入$ sho