我對PHP很新,我似乎無法找到我在找什麼。 我希望在填寫表單的頁面上顯示錯誤。 我想要一個「請輸入名稱」錯誤,以顯示它是否在表單頁面的$ showerror部分中爲空。窗體上的PHP表單錯誤消息頁面
這是我迄今爲止...
<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. [email protected]"></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>
那麼這就是我的process.php樣子......我不知道如何將錯誤代碼的一部分。
<?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 = '[email protected]';
$subject = 'Testing Form';
$body =
"Name: $name\n
E-Mail: $email\n
Multi Options: $multioption\n";
if ($submit && $human == '4') {
mail ($to, $subject, $body, $from);
print ("Thank you. We have received your inquiry.");
}
else {
echo "We have detected you are a robot!.";
}
?>
你確定你不是在尋找一個Javascript解決方案嗎? – Christoffer
這$淋浴錯誤是填充何時何地? – Aris
JavaScript更簡單,更高效嗎? – user2515606