如何将变量中存储的json数据从一页传递到php中的另一页?

时间:2018-12-31 07:13:02

标签: php jquery json ajax

我不知道这是否是重复的问题。 我想在不使用会话的情况下为登录表单编码。所以我在下面尝试了代码...

这是我的ajax电话

$('#flogin').submit(function(e){
   e.preventDefault();                
   var email = $('#email').val();
   var pass = $('#pass').val();
   var login = 
      'email='+email+
      '&pass='+pass;

   $.ajax({
     url : "data.php",
     type : "post",
     dataType : "json",
     data : login,
     success : function (disp){                        
         console.log(disp);
         var obj = disp;
         if(obj==""){
             alert("You entered wrong Email or Password...");
                 $('#flogin')[0].reset();
         }
          else { 
              location.href = "http://localhost/login/welcome.php";
          }                        
      }
   })
 })

ajax响应代码data.php

if (isset($_POST['email'])!="" && isset($_POST['pass'])!=""){
  $email = $_POST['email'];
  $pass = $_POST['pass'];
  $sql = " SELECT * FROM flogin WHERE email='$email' and pass='$pass' ";
  $result = mysqli_query($conn,$sql);
  $response = array();

  while ($menu = mysqli_fetch_assoc($result)){
    $response [] = array(
        'id'            => $menu['id'],
        'fname'         => $menu['fname'],
        'lname'         => $menu['lname'],
        'email'         => $menu['email'],
        'dob'           => $menu['dob']
    );
  }
  echo json_encode($response);
}

现在,我想将此数据发送(var obj = disp;)到'welcome.php'页面。但是我不知道如何将数据发送到该页面...

1 个答案:

答案 0 :(得分:0)

            <script type="text/javascript">
            $('.ownerround').click(function() 
            {
              var id=this.id;
               $.ajax({
                          type: 'POST',
                          url: '<?php echo site_url();?>Registration',
                          data: 'id='+id,
                            success: function()
                              {   
                         }); 
                              });

               </script>

               get id in model page:-

               $data=array('user_type'=>$this->input->post('id'));
               $this->db->insert('user',$data);
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