CODEWITHSUNDEEP
jqueryajaxcodeigniter
MGB C
MGB C

Reputation: 452

update text value to database with ajax in codeigniter

why it is my code doesn't work? maybe I miss something.

here is my code

View admin_page.php

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>

 <script type="text/javascript">

 // Ajax post
  $(document).ready(function() {

  $(".submit").click(function(event) {

  event.preventDefault();
  var message = $("input#l_message").val();

  jQuery.ajax({
  type: "POST",
  url: "<?php echo base_url(); ?>" + "admin/user_data_submit",
  dataType: 'json',
  data: {l_message: message},
   success: function(res) {
  if (res)
  {
   // Show Entered Value
   jQuery("div#msg").show();
    jQuery("div#msg").html(res..message);
  }
  }
  });

 });
 });
 </script>
<body>
<?php echo form_open();
    echo form_label('Librarians Message');?>
    <textarea class="form-control" name="the_librarian" id="l_message"></textarea>
    <?php echo form_submit('submit', 'Update',"class='submit'");?>
 <?php echo form_close();?>
</body>

in my controller admin.php

public function user_data_submit()
{
    $data = array('message' => $this->input->post('l_message'),
 );
  $this->home_admin_database->librarian_msg_insert($data);
  //Either you can print value or you can send value to database
  echo json_encode($data);
 }

then in my model Home_admin_database.php

public function librarian_msg_insert($data) {
$message = array(
           'message' => $data,
        );
 // Query to insert data in database
  $this->db->where('lm_id', '1');
  $this->db->update('librarians_message', $message);
  if ($this->db->affected_rows() > 0) {
     return true;
    }
   else {
   return false;
   }
  }

when I click the update button nothing happen and the page refresh without changing the database. help me please...

I want to update the database without refreshing the page then display a message after the success..

Upvotes: 1

Views: 2371

Answers (4)

Ahmad Faizal
Ahmad Faizal

Reputation: 21

Change your View code with this : 

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script>
 <script type="text/javascript">

  $(document).ready(function() {
        $("#submit").click(function(){
          // get all post from form with id form-msg
            dataString = $("#form-msg").serialize(); 
            $.ajax({
            type:"POST",
            url:"<?php echo base_url(); ?>index.php/admin/user_data_submit", 
            //parsing data to your controller
            data:dataString, 

            success:function (data) {
              // Show Entered Value
            $("div#msg").show();
            $("div#msg").html(data);
                alert('Success');
            }

            });

        return false;
        }); 
    }); 
 </script>

<body>

    <?php  
    // add form id
    echo form_open('',"id='form-msg'"); 
    echo form_label('Librarians Message');?>
    <textarea class="form-control" name="the_librarian" id="l_message"></textarea>    
    <?php 
    // add button
    echo form_button('submit', 'Update',"id='submit'"); 
    ?>
    <?php echo form_close();?>

</body>

Upvotes: 0

Freddy Sidauruk
Freddy Sidauruk

Reputation: 292

Hello i guess you using tinymce, tinymce cannot get value like that you should add to get value

tinyMCE.triggerSave();

then please change your javascript

 <script type="text/javascript">

 // Ajax post
  $(document).ready(function() {

  $(".submit").click(function(event) {

  event.preventDefault();
  tinyMCE.triggerSave();
  var message = $("#l_message").val();

alert(message) //please see is it right value or not !

  jQuery.ajax({
  type: "POST",
  url: "<?php echo base_url(); ?>" + "admin/user_data_submit",
  dataType: 'json',
  data: {l_message: message},
   success: function(res) {
  if (res)
  {
   // Show Entered Value
   jQuery("div#msg").show();
    jQuery("div#msg").html(res..message);
  }
  }
  });

 });
 });
 </script>

hope this help someone using tinymce,

Upvotes: 0

Ashish Raj
Ashish Raj

Reputation: 488

in your view file try using 

echo form_button('submit', 'Update',"class='submit'");

instead of 

echo form_submit('submit', 'Update',"class='submit'");

Try your luck with the submit event instead of click if above not working...

Also Replace

var message = $("input#l_message").val();

with

var message = $("#l_message").val();

=> Another step is to use 

<form Method="Post" class="submit_form">

instead of 

echo form_open();

or remove form action attribute by jquery on document load

and $(".submit_form").submit(function(event) {...}); for ajax call

Upvotes: 0

Praveen Kumar
Praveen Kumar

Reputation: 2408

Replace 

 var message = $("input#l_message").val();

with

var message = $("textarea#l_message").val();

Then it should work

Upvotes: 1

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