14

Thanks for taking the time to look, guys. I'm creating a pretty basic AJAX contact form using jQuery. The email sends, but upon opening the email there is no POST data, so I just get the strings I defined in the PHP script. On my phone's email client, the content of the email literally says 'undefined'. I've tried adding different types of header data to no avail, and a number of variations on the PHP mail() function.

I am more than willing to adopt an easier solution for a simple AJAX form, so thanks in advance for any new approaches.

Here is the form: 

   <section id="left">
      <label for="form_name">Name</label>
      <input name="form_name" id="form_name" type="text" >

      <label for="form_email">Email</label>
      <input name="form_email" id="form_email" type="email" >
   </section>

   <section id="right">
      <label for="form_msg">Message</label>
      <textarea name="form_msg" id="form_msg"></textarea>
      <input id="submit" class="button" name="submit" type="submit" value="Send">
   </section>

</form>

The jQuery AJAX:

$(function() {
    $("#contact .button").click(function() {
        var name = $("#form_name").val();
        var email = $("#form_email").val();
        var text = $("#msg_text").val();
        var dataString = 'name='+ name + '&email=' + email + '&text=' + text;

        $.ajax({
            type: "POST",
            url: "email.php",
            data: dataString,
            success: function(){
            $('.success').fadeIn(1000);
            }
        });

        return false;
    });
});

The PHP script (external file 'email.php'):

<?php
if($_POST){
    $name = $_POST['form_name'];
    $email = $_POST['form_email'];
    $message = $_POST['form_msg'];

//send email
    mail("email@gmail.com", "This is an email from:" .$email, $message);
}
?>
4

6 回答 6

22

无需创建查询字符串。只需将您的值放入一个对象中，jQuery 将为您处理其余的事情。

var data = {
    name: $("#form_name").val(),
    email: $("#form_email").val(),
    message: $("#msg_text").val()
};
$.ajax({
    type: "POST",
    url: "email.php",
    data: data,
    success: function(){
        $('.success').fadeIn(1000);
    }
});
于 2013-03-08T18:10:45.540 回答
6

保留您的 email.php 代码相同，但替换此 JavaScript 代码：

 var name = $("#form_name").val();
        var email = $("#form_email").val();
        var text = $("#msg_text").val();
        var dataString = 'name='+ name + '&email=' + email + '&text=' + text;

        $.ajax({
            type: "POST",
            url: "email.php",
            data: dataString,
            success: function(){
            $('.success').fadeIn(1000);
            }
        });

有了这个：

    $.ajax({
        type: "POST",
        url: "email.php",
        data: $(form).serialize(),
        success: function(){
        $('.success').fadeIn(1000);
        }
    });

以便您的表单输入名称匹配。

于 2013-03-08T18:34:33.900 回答
4

您使用了错误的帖子参数：

    var dataString = 'name='+ name + '&email=' + email + '&text=' + text;
                      ^^^^-$_POST['name']
                                       ^^^^--$_POST['name']
                                      etc....

javascript/html ID 与实际的 POST 无关，尤其是当您构建自己的数据字符串并且不使用相同的 ID 时。

于 2013-03-08T18:08:21.750 回答
3

您使用了错误的参数名称，请尝试：

if($_POST){
    $name = $_POST['name'];
    $email = $_POST['email'];
    $message = $_POST['text'];

//send email
    mail("j.andrew.sears@gmail.com", "51 Deep comment from" .$email, $message);
}
于 2013-03-08T18:14:54.783 回答
2

你的代码应该是：

   <section id="right">
      <label for="form_msg">Message</label>
      <textarea name="form_msg" id="#msg_text"></textarea>
      <input id="submit" class="button" name="submit" type="submit" value="Send">
   </section>

JS

var data = {
    name: $("#form_name").val(),
    email: $("#form_email").val(),
    message: $("#msg_text").val()
};
$.ajax({
    type: "POST",
    url: "email.php",
    data: data,
    success: function(){
        $('.success').fadeIn(1000);
    }
});

PHP：

<?php
if($_POST){
    $name = $_POST['name'];
    $email = $_POST['email'];
    $message = $_POST['text'];

//send email
    mail("email@gmail.com","My Subject:",$email,$message);
}
?>
于 2017-08-20T11:02:00.993 回答
0

您的 PHP 脚本（外部文件 'email.php'）应如下所示：

<?php
if($_POST){
    $name = $_POST['name'];
    $email = $_POST['email'];
    $message = $_POST['text'];

//send email
    mail("j.andrew.sears@gmail.com", "51 Deep comment from" .$email, $message);
}
?>
于 2013-03-08T18:11:04.357 回答