Question

我正在尝试将值传递给php文件。我收到提醒表单已提交，但没有显示在php文件中。我检查了js变量，它们正确显示。这是我的HTML表单。

                         <form name="sentMessage"role="form" data-toggle="validator" id="ContactForm" onsubmit="myFunction()">
                                <div class="form-group has-feedback">
                                    <input type="text" pattern="^[_A-z ]{1,}$" class="form-control" placeholder="Your Name *" name="name" id="Username"
                                    data-error="Plese enter your name" required>
                                    <span class="glyphicon form-control-feedback" aria-hidden="true"></span>
                                    <span class="help-block with-errors"></span>

                                </div>

                                <div class="form-group has-feedback">
                                    <input type="email" class="form-control" placeholder="Your Email *" name="email" id="Useremail" required data-validation-required-message="Please enter your email address.">
                                    <span class="glyphicon form-control-feedback" aria-hidden="true"></span>
                                    <span class="help-block with-errors"></span>

                                </div>
                                <div class="form-group has-feedback">
                                    <input type="tel" pattern="^[_0-9 ]{1,}$" class="form-control" placeholder="Your Phone *" name="phone" id="Userphone" required data-validation-required-message="Please enter your phone number.">
                                    <span class="glyphicon form-control-feedback" aria-hidden="true"></span>
                                    <span class="help-block with-errors"></span>
                                </div>
                                <div class="form-group">
                                    <textarea class="form-control" placeholder="Your Message *" name="message" id="Usermesage" required data-validation-required-message="Please enter a message."></textarea>

                                    <p class="help-block text-danger"></p>
                                    <label class="light">
                                </div>


                                <button type="submit" class="btn btn-xl btn-success">
                                    Send Message
                                </button>
                        </div>
                        </form>

 <script>
            function myFunction() {
                var name = document.getElementById("Username").value;
                var email = document.getElementById("Useremail").value;
                var phone = document.getElementById("Userphone").value;
                var message = document.getElementById("Usermesage").value;

                $.ajax({
                  type: 'POST',
                  url: 'mailer.php',
                  data: {'name': name, 'email':email,'phone':phone,'message':message},
                  success: function(){
                  alert("The form was submitted");
                }
                });


            }
    </script>

PHP文件mailer.php

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


}

?>

在php文件中没有显示任何内容。谢谢

Answer 1

两个想法 - 您是通过普通表单提交表单，然后尝试通过AJAX调用提交相同的表单。这样就不会给AJAX调用提供任何值，因为它们已经被提交了 - 你可能需要阻止正常的表单提交 - 这里有很多答案可以帮助你阻止表单提交直到AJAX函数。

第二 - 我甚至会从表单中删除提交按钮，给它一个type =“button”属性并组合onclick事件处理程序和ajax调用（请注意我使用的是serialize（）方法收集表单中的所有相关数据，而不是专门获取每个输入的值，而是： -

((lambda () (lambda (x) x)))

通过ajax将var传递给php？

1 个答案: