我的目标是使用php将表单数据作为电子邮件发送,表单div应该被另一个div替换。我已经使用jquery隐藏了div部分,但无法发送和发送电子邮件。我还编写了发送电子邮件的代码,但我的问题是如何调用有电子邮件发送代码的文件。
我的表单代码:
<form method="post" id="formsub">
<div id="form">
<div class="form-group">
<input type="text" name="name" class="form-control" id="name" placeholder="Name" required>
</div>
<div class="form-group">
<input type="text" name="email" class="form-control" id="email" placeholder="Email" required>
</div>
<div class="form-group">
<input type="text" name="phone" class="form-control" id="phone" placeholder="Phone Number" required>
</div>
<div class="form-group">
<input type="button" id="addbut" name="submit" value="Submit" class="form-control">
</div>
</div>
</form>
我的代码隐藏div并尝试过表单提交脚本:
<script>
$(document).ready(function() {
$("#addbut").on('click', function() {
$.ajax({
type: "POST",
url: "fromemail.php",
data: $(form).serialize(),
success: function(){
$("#form").hide();
$("#address").show();
}
});
});
});
</script>
我的php电子邮件发送代码:
<?php
if($_POST['submit']){
$to = "akhil@redd.xyz"; // this is your Email address
$from = $_POST['email']; // this is the sender's Email address
$name = $_POST['name'];
$phone = $_POST['phone'];
$subject = "Spots Contact";
$message = $first_name . ", with " . $phone . "has enquired for the service";
$headers = "From:" . $from;
mail($to,$subject,$message,$headers);
if(mail($to,$subject,$message,$headers))
{
echo "<script>alert('We will contact you shortly');</script>";
}
}
?>
答案 0 :(得分:0)
在表单操作属性中提供文件名:
<form id="formsub" method="post" action="fromemail.php">
并按照以下方式执行ajax代码:
$(document).ready(function(){
var form=$("#formsub");
$("#addbut").click(function(){
$.ajax({
type:"POST",
url:form.attr("action"),
data:$("#formsub").serialize(),
success: function(response){
console.log(response);
}
});
});
});
答案 1 :(得分:0)
@Rakhi ..
这是正确的吗?
<script type="text/javascript" src="assets/js/jquery-2.2.1.min.js"></script>
<script>
$(document).ready(function() {
var form=$("#formsub");
var base_url = "www.3ding.in/spots/";
$("#addbut").on('click', function() {
$("#form").hide();
$("#address").show();
$.ajax({
type: "POST",
url: base_url + "fromemail.php",
data: $("#formsub").serialize(),
success: function(response){
alert(1);
console.log(response);
}
});
});
});