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Home/ Questions/Q 8523449
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T07:24:45+00:00 2026-06-11T07:24:45+00:00

I have a view which has 2 forms: <table> <th>Write a comment.</th> <tr> <td>

  • 0

I have a view which has 2 forms:

<table>
<th>Write a comment.</th>
<tr>
    <td>
        <?php echo form_open($this->uri->uri_string(),$form1); 
              echo form_textarea($comment);
              echo form_submit('submit','submit');
              echo form_close();
        ?>
    </td>
</tr> 


</table>

<table>
    <tr>

        <td>
            <?php echo form_open($this->uri->uri_string()); 
                  echo form_dropdown('portion', $portion_options); 
                  echo form_submit('book','book');
                  echo form_close();
            ?>
        </td>
    </tr>
</table>

In the controller I check which button was clicked and then I perform some action by adding the corresponding form’s values to the database.

if(isset($_POST['book']))
{
    //sending the data to the database
    echo "Book button clicked";
}

if(isset($_POST['submit']))
{
   //sending the data to the database
   echo "Submit button clicked";
}

However when the ‘book’ button is clicked no action is performed. It is like the button was never clicked. Whereas when I click the ‘submit’ button, every action is done properly.

In the past I have used the same technique on plain php (i mean no framework, just php), and has worked fine for me. Does codeigniter need any further configuration? Am I doing something wrong?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-11T07:24:47+00:00Added an answer on June 11, 2026 at 7:24 am

    Why not add a hidden field to both forms called form_idwith values 1 and 2 respectively? Easy to catch in your controller upon post; e.g.:

    if($this->input->post()){
      switch($this->input->post('form_id')){
      case 1:
        // do stuff
      break;
      case 2:
        // do stuff
      break;
      }
    }
    
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