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Home/ Questions/Q 6151361
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T19:45:32+00:00 2026-05-23T19:45:32+00:00

$twoform = new TwoForm; $this->embedForm(‘twoform’, $twoform); if($value == false) { // unembedForm twoform HOW?

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$twoform = new TwoForm;

$this->embedForm('twoform', $twoform);

 if($value == false) {
  // unembedForm twoform  HOW?
}

how can i make if $value == false then form twoform is not submit and not add to database? unset fields not working, because form is sending, but it has a NULL value.

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  1. Editorial Team
    Editorial Team
    2026-05-23T19:45:33+00:00Added an answer on May 23, 2026 at 7:45 pm

    You can do this to unset form

    unset($this->widgetSchema['twoForm']);
    unset($this->validatorSchema['twoForm']);
    unset($this->formFieldSchema['twoForm']);

    But this won’t avoid posting the form in cas it was previously visible. You can replace validator to ignore posted values.
    $this->setValidator('twoForm', new sfValidatorPass());

    If the exemple given is complete, the better is imho not to embed the form if $value == false

    The database insertion depends on how your data is saved into it (sfDoctrineForm ?)

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