我创建了一个带有下拉列表值的联系表单,如下所示,但这些值未发送到我的电子邮件中-除下拉列表值外,其余所有值均通过发送。对此的任何帮助将不胜感激,因为我所做的研究似乎没有帮助。
Here is my JS
submitSuccess: function($form, event) {
event.preventDefault(); // prevent default submit behaviour
// get values from FORM
var name = $("input#name").val();
var email = $("input#email").val();
var radio = $("input#radio").val();
var message = $("textarea#message").val();
var firstName = name; // For Success/Failure Message
// Check for white space in name for Success/Fail message
if (firstName.indexOf(' ') >= 0) {
firstName = name.split(' ').slice(0, -1).join(' ');
}
$.ajax({
url: "assets/contact_me.php",
type: "POST",
data: {
name: name,
radio: radio,
email: email,
message: message,
submit: 1
},
Here is my PHP <?php if (isset($_POST['submit'])) {
var_dump($_POST);
$name=$_POST['name'];
$email=$_POST['email'];
$radio=$_POST['radio'];
$message=$_POST['message'];
$mailTo="info@margineco.com";
$subject="New message from ".$name;
$headers="From: ".$email;
$txt="You have received an email from ".$name." with regards to ".$radio.".\n\n".$message;
mail($mailTo, $subject, $txt, $headers);
header("Location: index.html?mailsend");
}
?>
Here is my HTML
<div class="row control-group">
<div class="form-group col-xs-12 controls">
<label>Plan Option<span>*</span></label>
<div class="dropdown">
<select class="form-control" id="radio" required data-validation-required-message="Please choose a option from the dropdown menu.">
<option type="radio" value="0" selected>Please select...</option>
<option type="radio" value="option1">General Inquiry</option>
<option type="radio" value="option2">Market Intelligence</option>
<option type="radio" value="option3">Economic Analysis</option>
<option type="radio" value="option4">Industry Forecasts</option>
</select>
</div>
<p class="help-block"></p>
</div>
</div>
答案 0 :(得分:0)
欢迎玛丽亚;)
第一:
<option type="radio" value="option4">Industry Forecasts</option>
“类型”实际上不是“选项”的有效选项:https://developer.mozilla.org/de/docs/Web/HTML/Element/option应该是:
<option value="option4">Industry Forecasts</option>
第二:
var radio = $("input#radio").val();
不会给您任何结果,因为您要从“选择”而不是“输入”中进行选择。应该是:
var radio = $("select#radio").val();