阿贾克斯笨

Question

我需要笨Ajax响应，请帮我做it..Please问任何其他进一步规定任何..

我已经得到空的响应，同时加入这种阿贾克斯 感谢你

function Ajax_newsletter() 
{ 

    var str=document.getElementById("sub_email").value; 

    if (str.length==0) 
     { 
     document.getElementById("txtHint").innerHTML=""; 

     } 
    if (window.XMLHttpRequest) 
     {// code for IE7+, Firefox, Chrome, Opera, Safari 
     xmlhttp=new XMLHttpRequest(); 
     } 
    else 
     {// code for IE6, IE5 
     xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); 
     } 
    xmlhttp.onreadystatechange=function() 
     { 
     if (xmlhttp.readyState==4 && xmlhttp.status==200) 
     { 
      alert(xmlhttp.responseText); 
      if(xmlhttp.responseText==1) 
      { 
      document.getElementById("txtHint").innerHTML='User Already Exist..'; 
      } 
      elseif(xmlhttp.responseText==0) 
      { 
       document.getElementById("txtHint").innerHTML='Newsletter subscribed..'; 
      } 
     } 
     } 
    xmlhttp.open("GET","<?php echo base_url();?>index.php/ajax/checkuser/"+str,true); 
    xmlhttp.send();  


} 

Answer 1

我假设你使用PHP，HTML，JavaScript（jQuery）。我会做这样的事情

在你看来将这个

$("#form-name").submit(function(e){ 
    e.preventDefault(); 
    dataString = $("#form-name").serialize(); 
    $.ajax({ 
     type:"POST", 
     url:"<?=base_url()?>controller_name/function_name.php", 
     data: dataString, 
     dataType: 'json', 
     success: function(data){ 
      if(data.response == "success"){ 
        alert("YAAAY"); 
      }else{ 
        alert("NOOOO..."); 
      } 
     } 
    }); 
}); 

在您的控制器确保你有这样的事情

public function function_name() 
{ 
    if($this->session->userdata('verified')) 
    { 
     $this->load->library('form_validation'); 
     $this->form_validation->set_rules('user-name', 'User Name', 'required|xss_clean'); 

     $this->form_validation->set_error_delimiters('<li>', '</li>'); 

     if($this->form_validation->run() == FALSE){ 
      $data['status'] = "failed"; 
      $data['message'] = "<h4>Failed to add new user, make sure all required information below is correct.<br/>"; 
      $data['message'] .= "<ul>"; 
      $data['message'] .= form_error('user-name'); 
      $data['message'] .= "</ul></h4>"; 
     }else{ 
      // WHAT SHOULD HAPPEN WHEN FORM IS VALID 
     } 
     echo json_encode($data); 
    }else{ 
      // WHAT SHOULD HAPPEN WHEN UNAUTHENTICATED USER IS ACCESSING THIS FUNCTION 
    } 
} 

Answer 2

正如你所标记jQuery试试这个：

function ajax_newsletter() { 
    var str = $("#sub_email").val(); 
    if (str.length == 0) { 
     $("#txtHint").html(""); 
    } 
    $.get("<?php echo base_url();?>index.php/ajax/checkuser/" + str) 
    .done(function(data) { 
     console.log(data); 
    }); 
}