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Jeirod avatar Drupal

Open new tab with external link using an image button

cn flag
Jeirod

I am making a custom module with a form that has an image button. The image button will open the external link with the value of a textfield inside the form. I have searched any possible solution in the internet but can't find a solution.

How do I code a submitForm button to open an external link with a form value of a textfield inside the form.

Below is my sample codes:

$form['search'] = [             
      '#type' => 'textfield', 
      '#title' => 'Search for',
      '#required' => TRUE
];

$form['submit'] = [
    '#type' => 'image_button',
    '#value' => $this->t(''),
    '#src' => 'modules/myModule/src/images/myImage.png',
    '#submit' => ['::submitForm'],
];

I also try to use these codes for the button but still no luck:

$form['submit'] = [
    '#type' => 'item',
    '#markup' => $this->t('
        <a href=\'https://www.google.com/search?%search\' target="_blank">
           <img width="30" height="30" border="0" align="center"  src=\'modules/myModule/src/images/myImage.png\'/>
        </a>', ['%search' => $search]),
];

Please note that these are not my actual codes but these are the concept of my codes.

Any comments and suggestions are welcome.

Thank you in advance.

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misterdidi avatar Drupal
de flag
misterdidi

If you use the formtarget HTML attribute on your image button, submit will happen in a new browser tab. Here is a basic example:

public function buildForm(array $form, FormStateInterface $form_state) {
  $form['search'] = [
    '#type' => 'textfield',
    '#title' => $this->t('Search for'),
    '#required' => TRUE,
  ];

  $form['image_button'] = [
    '#type' => 'image_button',
    '#value' => 'hello',
    '#src' => '/path/to/my/image',
    '#submit' => ['::mySubmitFunction'],
    '#attributes' => [
       // This attribute will open a new browser tab on form submission
      'formtarget' => '_blank' 
    ]
  ];

  return $form;
}

public function validateForm(array &$form, FormStateInterface $form_state) {
  // Do whatever form validation is needed
}

public function mySubmitFunction(array &$form, FormStateInterface $form_state) {
  $search = $form_state->getValue('search'); // Fetch 'search' field value
  // since we want to redirect to an external url, 
  // we need to use \Drupal\Core\Routing\TrustedRedirectResponse 
  // because redirect to external URL is not allowed by default
  $externalURL = \Drupal\Core\Url::fromUri('https://example.com', 
    ['query' => ['search' => $search]]
  );
  $response = new TrustedRedirectResponse($externalURL->toString());
  $form_state->setResponse($response);
}
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Jeirod avatar
cn flag
Jeirod
Thank you, @misterdidi. I have another question about your code. Is the `query` in `$externalURL` added to the https://example.com? So it will be https://example.com`$search`?
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misterdidi avatar
de flag
misterdidi
No, in my example, you'd end up with the URL https://example.com?search=my-search-value. If you want something like https://example.com/my-search-value you need to do something like `$externalURL = \Drupal\Core\Url::fromUri('https://example.com/' . $search);` (provided that you made sure that $search is url-friendly).
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Jeirod avatar
cn flag
Jeirod
Thanks, @misterdidi. Now I know how to make it happen. Thank you very much. It really helps me a lot.
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