如何使用javascript在PHP中发送单选按钮值？

How can I send radio button value in PHP using javascript?

本问题已经有最佳答案，请猛点这里访问。

我已经在这里呆了一个星期，读了所有的stackoverflow和google，我不能解决这个问题。每次我测试它，我都会得到第一个单选按钮，不管我点击哪个。

我收到的邮件是：

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Name: test
Email: test@live.com
phone: 9545027557
type: Residential

HTML：

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<form name="controls" id="controls" novalidate>

<input type="text" class="form-control"
placeholder="Full Name" id="name" required
data-validation-required-message="Please enter your name" />
<p class="help-block">
</p>

<input type="email" class="form-control" placeholder="Email"
id="email" required
data-validation-required-message="Please enter your email" />

<label class="btn btn-primary active">
<input type="radio" class="form-control" id="optionu" name="optionu"
value="Residential" checked > Residential
</label>
<label class="btn btn-primary">
<input type="radio" class="form-control" id="optionu" name="optionu"
value="Commercial"> Commercial
</label>
<label class="btn btn-primary">
<input type="radio" class="form-control" id="optionu" name="optionu"
value="Handyman"> Handyman
</label>

<p>

</p>


<button type="submit" class="btn btn-danger pull-right">Request</button><br />
</form>

号

联系u me.js：

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var name = $("input#name").val();
var email = $("input#email").val();
var phone = $("input#phone").val();
var optionu = $("input#optionu").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:"bin/contact_me.php",
type:"POST",
data: {name: name, email: email, optionu: optionu, phone: phone},
cache: false,
success: function() {
// Success message
$('#success').html("");
$('#success > .alert-success').html("<button type='button' class='close' data-dismiss='alert' aria-hidden='true'>×")
.append("</button>");
$('#success > .alert-success')
.append("Your message has been sent.");
$('#success > .alert-success')
.append('');

联系u me.php：

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$name = $_POST['name'];
$optionu_value = $_POST['optionu'];
$phone = $_POST['phone'];
$email_address = $_POST['email'];

// create email body and send it
$to = '[email protected]'; // put your email
$email_subject ="Contact form submitted by: $name";
$email_body ="You have received a new message.

".
" Here are the details:

Name: $name
".
"Email: $email_address
phone: $phone
type: $optionu";

。

相关讨论

首先，你使用同一个ID 4次。ID在文档中必须是唯一的，不能将同一ID设置为4个元素。我也认为你从单选按钮中获取价值的方式是错误的。

这或多或少就是如何从单选按钮中获取值的方法：(另一个so问题的例子)

1	var optionu = $('input[name=optionu]:checked', '#controls').val()

同时从所有输入[radio]元素中删除id。

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<label class="btn btn-primary active">
<input type="radio" class="form-control" name="optionu" value="Residential" checked /> Residential
</label>
<label class="btn btn-primary">
<input type="radio" class="form-control" name="optionu" value="Commercial" /> Commercial
</label>
<label class="btn btn-primary">
<input type="radio" class="form-control" name="optionu" value="Handyman" /> Handyman
</label>

号