TL; DR解决方案:将javascript中的.val更改为.serialize以获取任何无线电输入。
我一直在使用this tutorial来构建一个表单,当按下提交按钮时,淡出按钮并淡入“谢谢”消息并在后台发送mailer.php。我的表单有单选按钮,我似乎无法弄清楚如何通过我的电子邮件将javascript发送到哪个按钮。
这是表单html:
<form action="" method="" name="rsvp" id="rsvp-form">
<fieldset>
<legend>RSVP</legend>
<ol>
<li>
<input id="accepts1" class="rsvps" name="rsvps" type="radio" value="Graciously_Accepts" />
<label for="accepts1">Graciously Accepts</label>
</li>
<li>
<input id="declines1" class="rsvps" name="rsvps" type="radio" value="Regretfully_Declines" />
<label for="declines1">Regretfully Declines</label>
</li>
<li>
<input id="accepts2" class="rsvps" name="rsvps" type="radio" value="Regretfully_Accepts" />
<label for="accepts2">Regretfully Accepts</label>
</li>
<li>
<input id="declines2" class="rsvps" name="rsvps" type="radio" value="Graciously_Declines" />
<label for="declines2">Graciously Declines</label>
</li>
</ol>
</fieldset>
<div id="rsvp-wrapper">
<fieldset>
<button class="button" type="submit" value="send">RSVP!</button>
</fieldset>
</form>
<div class="success"></div>
</div>
javascript:
<script type="text/javascript">
$(function() {
$(".button").click(function() {
var rsvps = $(".rsvps").val();
var dataString = 'rsvps=' + rsvps;
$.ajax({
type: "POST",
url: "rsvp-mailer.php",
data: dataString,
success: function() {
$('#rsvp-wrapper').html("<div class='success'></div>");
$('.success').html("<p class='italic'>Thanks!</p>")
.hide()
.fadeIn(500, function() {
$('.success');
});
}
});
return false;
});
});
</script>
和mailer.php:
<?php
$rsvps = $_POST['rsvps'];
$formcontent="
RSVP: $rsvps \n";
$recipient = "myemail@domain.com";
$subject = "RSVP";
$mailheader = "RSVP \r\n";
mail($recipient, $subject, $formcontent, $mailheader) or die("Error!");
?>
非常感谢您提供的任何见解。
答案 0 :(得分:4)
试一试。有关详细信息,请参阅jQuery.post()。
<script type="text/javascript">
$('form').submit(function() {
var data = $(this).serialize();
$.post("rsvp-mailer.php", data, function() {
$('#rsvp-wrapper').html("<div class='success'></div>");
$('.success').html("<p class='italic'>Thanks!</p>")
.hide()
.fadeIn(500, function() {
$('.success');
});
}
return false;
}
</script>
答案 1 :(得分:1)
不要通过类选择器访问单选按钮,而是尝试以下操作:
var rsvps = $('input[name=rsvps]:radio').val();
答案 2 :(得分:0)
将var rsvps = $(".rsvps").val();
更改为var rsvps = $(".rsvps[selected=selected]").val();
此外,dataString需要是一个像这个var dataString = { rsvps : rsvps };
的json对象,可以通过ajax POST访问。