我的javascript和php目前非常低,我正在尝试制作一些非常简单但只能做到的东西!
我有一个表单,我希望将checbox值(标记为有效的那些)传递给他们需要访问的文件(除了复选框之外它们都很好并传递)并从那里到电子邮件(除了复选框之外,它也有效。)
这是我的代码:
(再次 - 一切正常,但复选框和我删除了一些小东西,如其他东西的验证,所以它将是可以阅读的)
<script type="text/javascript">
jQuery(function () {
jQuery(".button").click(function () {
// validate and process form
// first hide any error messages
jQuery('.error').hide();
var name = jQuery("input#name").val();
var email = jQuery("input#email").val();
var phone = jQuery("input#phone").val();
var dataString = 'name=' + name + '&email=' + email + '&phone=' + phone;
//alert (dataString);return false;
jQuery.ajax({
type: "POST",
url: "mitkadmim.php",
data: dataString,
success: function () {
jQuery('#contact_form').html("<div id='message'></div>");
jQuery('#message').html("<h2>Contact Form Submitted!</h2>")
.append("<p>We will be in touch soon.</p>")
.hide()
.fadeIn(1500, function () {
jQuery('#message').append("<img id='checkmark' src='images/check.png' />");
});
}
});
return false;
});
});
</script>
<div id="mheo">
<div id="contact_form">
<form name="contact" method="post" action="">
<fieldset>
<label for="name" id="name_label">שם: </label>
<input type="text" name="name" id="name" size="30" value="" class="text-input" /><br/><br/>
<label for="email" id="email_label">E-mail: </label>
<input type="text" name="email" id="email" size="30" value="" class="text-input" /><br/><br/>
<label for="phone" id="phone_label">מס. טלפון</label>
<input type="text" name="phone" id="phone" size="30" value="" class="text-input" /><br/><br/>
intrested in:<br/>
<input type="checkbox" name="kidumchoise[]" value="kidum_esek" />aaa<br />
<input type="checkbox" name="kidumchoise[]" value="mitug_esek" />bbb<br />
<input type="checkbox" name="kidumchoise[]" value="laikim" />CCC<br />
<input type="checkbox" name="kidumchoise[]" value="aher" />DDD<br />
<br />
<input type="submit" name="submit" class="button" id="submit_btn" value="Send" />
</fieldset>
</form>
</div>
</div>
和电子邮件端
<?php
$recipient = 'ddsds@gmail.com';
$name = $_POST['name'];
$email = $_POST['email'];
$phone = $_POST['phone'];
$kidumchoise = $_POST['kidumchoise'];
$msg = "Message from: $name\n\n".$_POST['phone']."\n".$_POST['email'].$_POST['kidumchoise']." \n""\n".$_POST['kidumchoise']."\n";
mail($recipient, $name , $msg);
?>
我做错了什么? 谢谢,
答案 0 :(得分:4)
var chkElem = document.getElementsByName("kidumchoise[]");
var choice ="";
for(var i=0; i< chkElem.length; i++)
{
if(chkElem[i].checked)
choice = choice + chkElem[i].value;
}
var dataString = 'name='+ name + '&email=' + email + '&phone=' + phone + '&kidumchoise=' +choice;
在以下行
之后添加以下代码 var phone = jQuery("input#phone").val();
它会正常工作。