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Home/ Questions/Q 8865387
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T16:29:36+00:00 2026-06-14T16:29:36+00:00

So I’m trying to output a session’s variable value into a view (only if

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So I’m trying to output a session’s variable value into a view (only if it exists). But I can’t seem to get it to work. No error. Nothing. What am I missing? Thought I had this down packed by now – guess not…

In controller…FYI, $data is assigned to the view (i.e. $this -> load -> view('view', $data);

$data['campaign_name'] = $this -> session -> userdata('campaign_name');

Here’s my php snippet in the view that I’m trying to output. So in short, if the session exists, output it. If not, do nothing.

<input type="text" name="campaign_name" class="wizardInput nameField" value="<? if (isset($campaign_name)) ;?> ">

Anyone?

EDIT Okay, I should have mentioned that i’m trying to output the session value into a FORM value. Modified view code above. The form submits as though the value is there – and even sends the value along. However, it’s not visible in the text input…

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  1. Editorial Team
    Editorial Team
    2026-06-14T16:29:40+00:00Added an answer on June 14, 2026 at 4:29 pm

    you can easly do this in your view:

    if($this->session->userdata('campaign_name')){
     // do somenthing cause it exist
    }
    

    then if you want to make session data as the value of an input do this:

    <input type="text" name="campaign_name" class="wizardInput nameField" value="<?php echo $this->session->userdata('campaign_name') ?>">
    

    you don’t need to control if session userdata exist cause if it not exist it doesn’t prints anything, cause userdata() method returns false !

    Then you don’t need to pass session data trough the $data[] array, cause session data can be retrieved from anywhere (model/controllers/views/hooks/libraries/helpers and so on)

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