Question

我是PHP的新手，我似乎无法找到我想要的东西。我希望错误显示在填写表单的页面上。我想要一个“请输入名称”错误，以显示表单页面的$ showerror部分中是否为空。

这就是我到目前为止......

<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>

然后这就是我的process.php看起来的样子......我不知道如何编写错误部分。

<?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，此行：

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

变成这样：

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

如果您对PHP完全陌生，可以从一些教程网站开始学习，这里是good one

修改

您可以在PHP页面中设置$ showerror，这可能是每个表单字段的“if conditions”的长列表，但我将向您展示全名$_POST['name']的一个小/简单示例，它会是这样的：

$showerror = ''; if(!empty($_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

编译器只解析内容为php，它写在<?php ?>标记之间，所以使用

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

或

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

Answer 3

试试这个，

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

如果您在第一次编辑页面时没有验证Undefined variable变量的代码，则会收到错误消息$showerror

Answer 4

试试这个..你必须将两个代码都放在同一个文件中。并且可以检查请求是否发布只读了php。然后你可以将值放在$ sho

中

表单页面上的PHP表单错误消息

4 个答案: