JQuery Ajax POST in Codeigniter

问题:

I have searched a lot of tutorials with POST methods and saw answered questions here too but my POST still doesn't work...I thought i should post it here if you guys see something that i don't!

My js - messages.js:

$(document).ready(function(){   

    $("#send").click(function()
    {       
     $.ajax({
         type: "POST",
         url: base_url + "chat/post_action", 
         data: {textbox: $("#textbox").val()},
         dataType: "text",  
         cache:false,
         success: 
              function(data){
                alert(data);  //as a debugging message.
              }

     return false;
 });
 });

My view - chat.php:

<?php $this->load->js(base_url().'themes/chat/js/messages.js');?> //i use mainframe framework which loading script this way is valid



<form method="post">
    <input id="textbox" type="text" name="textbox">
    <input id="send" type="submit" name="send" value="Send">
</form>

Last My controller - chat.php

//more functions here

function post_action()
{   
    if($_POST['textbox'] == "")
    {
        $message = "You can't send empty text";
    }
    else
    {
        $message = $_POST['textbox'];
    }
    echo $message;
}

回答1:

$(document).ready(function(){   

    $("#send").click(function()
    {       
     $.ajax({
         type: "POST",
         url: base_url + "chat/post_action", 
         data: {textbox: $("#textbox").val()},
         dataType: "text",  
         cache:false,
         success: 
              function(data){
                alert(data);  //as a debugging message.
              }
          });// you have missed this bracket
     return false;
 });
 });

回答2:

In codeigniter there is no need to sennd "data" in ajax post method.. here is the example .

   searchThis = 'This text will be search';
    $.ajax({
      type: "POST",
      url: "<?php echo site_url();?>/software/search/"+searchThis,
      dataType: "html",
      success:function(data){
        alert(data);
      },

    });

Note : in url , software is the controller name , search is the function name and searchThis is the variable that i m sending.

Here is the controller.

    public function search(){
    $search = $this->uri->segment(3);
      echo '<p>'.$search.'</p>';
    }

I hope you can get idea for your work .

回答3:

    <?php if ( ! defined('BASEPATH')) exit('No direct script access allowed');

    class UserController extends CI_Controller {

        public function verifyUser()    {
            $userName =  $_POST['userName'];
            $userPassword =  $_POST['userPassword'];
            $status = array("STATUS"=>"false");
            if($userName=='admin' && $userPassword=='admin'){
                $status = array("STATUS"=>"true");  
            }
            echo json_encode ($status) ;    
        }
    }


function makeAjaxCall(){
    $.ajax({
        type: "post",
        url: "http://localhost/CodeIgnitorTutorial/index.php/usercontroller/verifyUser",
        cache: false,               
        data: $('#userForm').serialize(),
        success: function(json){                        
        try{        
            var obj = jQuery.parseJSON(json);
            alert( obj['STATUS']);


        }catch(e) {     
            alert('Exception while request..');
        }       
        },
        error: function(){                      
            alert('Error while request..');
        }
 });
}

回答4:

The question has already been answered but I thought I would also let you know that rather than using the native PHP $_POST I reccomend you use the CodeIgniter input class so your controller code would be

function post_action()
{   
    if($this->input->post('textbox') == "")
    {
        $message = "You can't send empty text";
    }
    else
    {
        $message = $this->input->post('textbox');
    }
    echo $message;
}