CodeIgniter - 2 forms in a view

Question

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?

You should be able to have 2 forms on a page just fine. Can you post the uri_string that you're hitting in your browser and then post your entire method? — Catfish
– Catfish, Commented Sep 13, 2012 at 14:54
I do have it on the top of my view. This is the array: $form1 = array( 'name' => 'form1', 'id' => 'form1' ); — Hakuna Matata
– Hakuna Matata, Commented Sep 13, 2012 at 14:54
Sorry for my delay. I was looking whether the problem is on the dropdown, and maybe it is there, since I changed the input from a dropdown to a text input and now the clicked button is beeing recognized — Hakuna Matata
– Hakuna Matata, Commented Sep 13, 2012 at 15:04

Mudshark · Accepted Answer · 2012-09-13 14:13:36Z

1

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;
  }
}

answered Sep 13, 2012 at 14:13

Mudshark

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6 Comments

Hakuna Matata Over a year ago

Thanks for your reply. I just tried it your way, however, still just the 'submit' button is recognized

Mudshark Over a year ago

Are you using if($this->input->post('book')) {or raw if(isset($_POST['book'])) { ?

Hakuna Matata Over a year ago

Actually I used the $this->input->post() since I thought that from the swich is going to be understood which form was posted. I tried now with $this->input->post('book') and still nothing.

Mudshark Over a year ago

I just tested locally, having echo form_hidden('form_id', 1); and echo form_hidden('form_id', 2); on both forms respectively. When in the controller I do if($this->input->post()){ echo $this->input->post('form_id'); } I get correct output - so it should work.

Mudshark Over a year ago

I think I figured it out. You're checking for if(isset($_POST['submit'])) Since both buttons are submit buttons, your second ifwill always evaluate as true. So you need to check not for the existence of a submit but for the existence of another element, the hidden field as I suggested. Or, you could do this: if($_POST['submit'] == 'submit'){ and if($_POST['submit'] == 'book'){

|

Sergey Telshevsky · Accepted Answer · 2012-09-13 13:54:17Z

0

<?php echo form_open($this->uri->uri_string(),$form1);

and

<?php echo form_open($this->uri->uri_string());

Looks like you forgot to provide the settings in the second one like:

<?php echo form_open($this->uri->uri_string(),$form2);

answered Sep 13, 2012 at 13:54

Sergey Telshevsky

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6 Comments

Hakuna Matata Over a year ago

no, either with <?php echo form_open($this->uri->uri_string()); or <?php echo form_open($this->uri->uri_string(),$form2); does not work :/ $form1 = array( 'name' => 'form1', 'id' => 'form1' ); and $form2 = array( 'name' => 'form2', 'id' => 'form2' );

Hakuna Matata Over a year ago

When I unite the forms to 1, then which button was clicked is recognized and work properly. However I need them to be on seperate forms if this is possible.

Sergey Telshevsky Over a year ago

In any case do not use if(isset($_POST['book'])) in CodeIgniter. Use built in $this->input->post('book');

Sergey Telshevsky Over a year ago

And see in firebug/chrome developer tools what data is being sent to the controller, it can be seen in Network tab.

Hakuna Matata Over a year ago

Even though I do not know how to use the firebug, in the Headers Tab at the Network tab I see that what I selected from the dropdown has been sent

|

Hakuna Matata · Accepted Answer · 2012-09-13 21:37:30Z

Well after spending my whole day on this I finaly managed to solve it somehow (although I believe it is not such a propper way to handle this).

Well:

$comment = array(
    'name'      => 'comment',
    'id'        => 'comment',
    'value'     => 'write you comment',
    'row'       => '5',
    'cols'      => '100'
    );

<table>
<th>Write a comment.</th>
<tr>
    <td>
        <?php echo form_open($this->uri->uri_string()); 
              echo form_hidden('form_id', 1);
              echo form_textarea($comment);
              echo form_submit('submit','submit');
              echo form_close();
        ?>
    </td>
</tr> 


</table> 

<table>
    <th>Write a comment.</th>
    <tr>
        <td>

            <?php echo form_open($this->uri->uri_string()); 
                  echo form_hidden('form_id', 2);
                  echo form_dropdown('comment', $portion_options);
                  echo form_submit('book','book');
                  echo form_close();
            ?>

        </td>
    </tr> 


</table>

Probably the forms fields (textarea and dropdown) needed to have the same name (which i have set to 'comment'). Although I do not understand why :/

Thank you all for trying to help me :)

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