Codeigniter displaying ajax request

Question

I'm a Codeigniter noob. I've been trying for quite some time now to solve this. I have a textview in my view.php. On button press, I want to send the text to server (php file) , process it and display result on page. My current code is:

javascript in view:

function verify(){
                var posttext = $('#post_text').text();
                $.ajax({
                    type: "post",
                    url: "http://localhost/naloga1/CodeIgniter/index.php/usercontroller/checkinput",
                    cache: false,               
                    data: {post_text : posttext },
                    success: function(json){                        
                    try{        
                        var obj = jQuery.parseJSON(json);
                        alert( obj['STATUS']);


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

usercontroller

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

class UserController extends CI_Controller {


    public function checkinput(){
        $status = array("STATUS"=>"false");
        $text =  $_POST['post_text'];
        if ($text != null){
            $status = array("STATUS"=>"true");  
        }
        echo json_encode ($status) ;
        $output = $text.strip_tags();
        echo "<p>".$output."</p>";
    }   
}

and my textview

<textarea rows="3" cols="25" id="post_text" >Some random text</textarea>
    <input type="submit"  id="post_bttn" value="Post" onclick="verify()">

Any help would be greatly appreciated.

echo json_encode ($status) ; $output = $text.strip_tags(); echo "<p>".$output."</p>"; why are you echo two times — shafiq.rst
– shafiq.rst, Commented Aug 21, 2015 at 6:37

user3232286 · Accepted Answer · 2015-08-21 06:31:42Z

2

function verify(){
        var posttext = $('#post_text').text();
        $.ajax({
            type: "post",
            url: "http://localhost/naloga1/CodeIgniter/index.php/usercontroller/checkinput",
            cache: false,               
            data: {post_text : posttext },
            success: function(json){                        
            try{        
                var obj = jQuery.parseJSON(json);
                alert( obj['STATUS']);


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

answered Aug 21, 2015 at 6:31

user3232286

Sign up to request clarification or add additional context in comments.

3 Comments

sdgaw erzswer Over a year ago

I tried it and it throws me the Exception while request error, am I missing something?

sdgaw erzswer Over a year ago

I dont't get any error apart from this javascript alert

user3232286 Over a year ago

$output = $text.strip_tags(); echo "<p>".$output."</p>"; this thing is not encode in json, so you need to encode it and try to keep all in one echo only encode everything and only echo for one time

Saty · Accepted Answer · 2015-08-21 06:42:48Z

0

Right now you are not passing any parameter to your Ajax request. Your Data would be

 data: {post_text:$('#post_text').text()},

And in controller file Use strip_tags() like below

 $output = strip_tags($text,'<br><br/>');

edited Aug 21, 2015 at 6:42

answered Aug 21, 2015 at 6:30

Saty

22.5k7 gold badges35 silver badges52 bronze badges

9 Comments

sdgaw erzswer Over a year ago

Thank you, apparently this was my first mistake, but it still throws me an error..

sdgaw erzswer Over a year ago

alert('Exception while request..'); This part of js appears (a window with this text)

sdgaw erzswer Over a year ago

Thank you, but it still throws me the same exception. This is weird.

Saty Over a year ago

Now problem is with try var obj = jQuery.parseJSON(json); alert( obj['STATUS']); just alert(json); on it and check

Saty Over a year ago

try{ alert(json); } and check the result

|

Collectives™ on Stack Overflow

Codeigniter displaying ajax request

2 Answers 2

3 Comments

9 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

9 Comments

Your Answer

Sign up or log in

Post as a guest

Related