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Home/ Questions/Q 5981919
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Editorial Team
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Editorial Team
Asked: May 22, 20262026-05-22T22:00:45+00:00 2026-05-22T22:00:45+00:00

I’m loading javascript files in the bootstrap as usual, but there’s a file that

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I’m loading javascript files in the bootstrap as usual, but there’s a file that I want to have included only if it’s a page that has a form

->appendFile('http://myurl.com/js/formscript.js');

Is there a way to detect the page being loaded, from the bootstrap so I can decide whether or not to include this file?

I thought about passing a variable from the Form to the view, and then checking for that variable in the bootstrap, but it’s not working.

This would be in my form

$layout = new Zend_Layout();
$view = $layout->getView();     
$view->formscript = true; 

and this would be in my bootstrap

if ($view->formscript)

but var_dump($view->formscript) give me null, so any other ideas to activate js files only in specific conditions?

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  1. Editorial Team
    Editorial Team
    2026-05-22T22:00:46+00:00Added an answer on May 22, 2026 at 10:00 pm

    It is possible, but you do not need your bootstrap. You can just access the variable from your layout:

    //form
    $view = Zend_Layout::getMvcInstance()->getView();
    $view->formscript = TRUE;
    
    //layout
    if($this->formscript)
    {
      $this->headScript()->appendFile('http://myurl.com/js/formscript.js');
    }
    echo $this->headScript();
    

    Do not use getView() in your form as it will return the view object for the form, not for your application. This has tripped me up more than a couple of times >.>

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