Generating Javascript view block with CakePHP

Question

Consider a URL like this:

http://site.com/upload/qeSJGs,ZWURb4,qdMMTZ,yM62UX,RlwwWT,ecw7s1

I need to get all of the 6 character strings separated by , into a Javascript array and inserted into my page. Here's my controller action:

public function upload($imageHashes) {
    $this->set('title_for_layout', 'Uploads successful');
    $this->set('imageHashes', explode(',', $imageHashes);
    $this->layout = 'complex';
}

and in my view file I have this:

<?php echo $this->Html->scriptBlock('', array('inline' => false)); ?>

Now, it's going to be mighty messy to write all the Javascript in the scriptBlock method, but I can't include an external Javascript file because the content changes based on the URL. Is there an easier way to do this that I'm not aware of?

Mike Brant · Accepted Answer · 2012-07-18 21:29:30Z

4

Take your imageHashes array and json_encode it. Then for your script block you can just do something like

echo $this->Html->scriptBlock('var jsArray = ' . $json_encoded_array . ';', array('inline' => false));

answered Jul 18, 2012 at 21:29

Mike Brant

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1 Comment

Mike Brant Over a year ago

No problem. Was this the answer you were looking for?

Iurii Tkachenko · Accepted Answer · 2016-06-20 17:45:15Z

3

For Cakephp 2.x and Cakephp 3.x You can use:

In View or Template file:

<?php $this->Html->scriptStart(array('block' => 'scriptBottom', 'inline' => false)); ?>

$(document).ready(function(){
  console.log('OK');
});

<?php $this->Html->scriptEnd(); ?>

In Layout file:

<?= $this->fetch('scriptBottom'); ?>

edited Jun 20, 2016 at 17:45

Iurii Tkachenko

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answered Jun 20, 2016 at 17:01

Konya Istvan

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