Question

所以我像Ajax一样提交我的表单

$("#submitform").click(function(e){
e.preventDefault();
var form_data = $("#contactfrm").serialize();
$.ajax({
    type: "POST",
    url: "/ltlcreation-new/wordpress/wp-content/themes/LTLCreation/includes/form-handler.php",
    data: form_data,
    error: function(){
        alert("failed");
    },
    success: function(json_data){
        console.log(json_data);
        alert("success");
    },
    })
});

在我的form-handler.php中，我捕获了来自错误

    <?php


if(isset($_POST['submit'])) {

  //include validation class
  include 'validate.class.php';

  //assign post data to variables
  $name = @($_POST['name']);
  $email = @($_POST['email']);
  $message = @($_POST['message']);
  $phone = @($_POST["phone"]);

  //echo $name, $email, $message, $phone;


  //start validating our form
  $v = new validate();
  $v->validateStr($name, "name", 3, 75);
  $v->validateEmail($email, "email");
  $v->validateStr($message, "message", 5, 1000);
  $v->validateStr($phone, "phone", 11, 13);

  if(!$v->hasErrors()) {

        $to = "lukelangfield001@googlemail.com";
        $subject = "Website contact form ";

        $mailbody = $message . "\n" . "from " . $name . "\n" . $phone;
        $headers = "From: $email";

        mail($to, $subject, $mailbody, $headers);

        echo "success";

    } else {
    //set the number of errors message
    $message_text = $v->errorNumMessage();

    //store the errors list in a variable
    $errors = $v->displayErrors();

    //get the individual error messages
    //$nameErr = $v->getError("name");
    //$emailErr = $v->getError("email");
    //$messageErr = $v->getError("message");
    //$phoneErr = $v->getError("phone");

   echo $message_text;  echo $errors;  

   $output = array($message_text);
   echo json_encode($output);
  }//end error check

}// end isset

这些错误通常看起来像这样

There were 4 errors sending your message!

Name must be at least 3 characters long.
Please enter an Email Address.
Message must be at least 5 characters long.
Phone must be at least 11 characters long.

["There were 4 errors sending your message!\n"]

我已尝试对输出进行jSon编码，并在ajax中成功将json数据拉出，但它只是不断返回一个空字符串

（空字符串）

我的问题是你可以将数据从PHP发送回Ajax，如果是这样的话我完全错了吗？

Answer 1

您可以输出除json字符串以外的任何内容，以便echo "success";生成t。使用调试器网络响应输出选项卡可以看到这是正确编码的。

也不要使用

$name = @($_POST['name']);

改为使用

$name = isset($_POST['name']) ? $_POST['name'] : '';

如果您仍有空白页，请确保设置了显示错误。

error_reporting(E_ALL);
ini_set('display_errors', 1);

Answer 2

我是一个白痴我仍然在我的PHP文件中有这个，这意味着表单没有触发或返回响应，愚蠢的我，很高兴我终于想通了

if(isset($_POST['submit'])) {

感谢帮助人员

Answer 3

以下是您可以使用的Ajax联系表单示例：

Ajax.js

$(document).ready(function(){
        $("#btn").click(function(){

                var username=$("#name").val();
                var email=$("#email").val();
                var dis=$("#dis").val();
                var process=true;
                if(username=="")
                    process=false;
                if(email=="")
                    process=false;
                if(dis=="")
                    process=false;

                if(process){

                var dataString="name="+username + "&email="+email+ "&message="+dis;
                    $("#res").html('<span>Sending...</span><img src="a.gif">'); 
                    $.ajax({
                        url:"b.php",
                        type:"POST",
                        data:dataString,
                        success:function(data){
                            document.getElementById("name").value='';
                            document.getElementById("email").value='';
                            document.getElementById("dis").value='';
                            $("#res").html(data);
                        }
                    });
                }else{
                    alert("fill all fields");
                }   
            });

    });

和b.php

<?php
mysql_connect("localhost","root","");
mysql_select_db("ajax") || die("erro");

if(isset($_POST['name'])){
     mysql_real_escape_string(htmlentities($name=$_POST['name']));
     mysql_real_escape_string(htmlentities($email=$_POST['email']));
     mysql_real_escape_string(htmlentities($message=$_POST['message']));
     if(!empty($name) && !empty($email) && !empty($message)){
    if(mysql_query("INSERT INTO `users` (name,email,message) VALUES('$name','$email','$message') ")){

        echo 'The massage has been send';
    }else{
        echo mysql_error();
        }
    }


}
?>

享受....

Answer 4

您有以下内容：

success: function(json_data){

虽然json_data根本没什么。它应该是

success: function(data){

将数据从php发送回Ajax

4 个答案: