为什么我没有从 php 得到任何东西？

Question

$('#registerForm').submit(
    function()
    {
        callAjaxSubmitMethod(this);
        //event.preventDefault();
    }
);

是我里面的函数dom准备好了。

function callAjaxSubmitMethod(form)
{
    $.ajax({
        type: "POST",
        url: "lib/registration_validate.php",
        data: $("#registerForm").serialize(),

        success: function(response)
        {
            alert("s"+response.status);
        },

        error:function(response)
        {
            alert("e"+response);
        }
    });
}

是实际的函数定义。

我的php内容是

<?php
include 'configdb.php';
session_start();
global $connection;
echo "oops";
if(isset($_POST['submit']) && $_POST['email']!='' && $_POST['password']!='')
{   //use empty....
   $email= $_POST['email'];
   $sql1 = "SELECT * FROM Users WHERE emailID = '$email'";
   $result1 = mysqli_query($connection,$sql1) or die("Oops");
   if (mysqli_num_rows($result1) > 0)
   {
        $_SESSION['email_exist']="Email Already Exists";
        //die({'status':'Email Already Exists'});
        //echo var_dump($_SESSION);
        //echo '{status:0}' ;
        //exit;
   }
  }
  ?>

我没有收到任何提醒。如果我启用 event.preventDefault() 我在 mozilla 中得到 out of memory error 并且在 chrome 中得到 sundefined 警报。

Answer 1

使用

if(isset($_POST['submit']) && $_POST['email']!='' && $_POST['password']!='')
{   //use empty....
   $email= $_POST['email'];
   $sql1 = "SELECT * FROM Users WHERE emailID = '$email'";
   $result1 = mysqli_query($connection,$sql1) or die("Oops");

   $response = array();

   if (mysqli_num_rows($result1) > 0)
   {
        $_SESSION['email_exist']="Email Already Exists";

        $response['status']='Email Already Exists';
   }
   else{
        $response['status']='Allis OK';
   }
   echo json_encode($response); die;
}

并在 ajax 中添加：

dataType:'json'

编辑

关于如何设置数据类型的信息：

$.ajax({
        type: "POST",
        dataType:'json',
        url: "lib/registration_validate.php",
        data: $("#registerForm").serialize(),

        success: function(response)
        {
            alert("s"+response.status);
        },

        error:function(response)
        {
            alert("e"+response);
        }
    });

为什么我没有从 php 得到任何东西？

Why am i not getting anything from php?

php

ajax

local-storage