Symfony2: How to get asset's URL in function inside view template

Question

As documented here I can read my image files in view by using following code.

<img src="<?php echo $view['assets']->getUrl('images/logo.png') ?>" alt="Symfony!" />
////  This outputs the image

or I can also do

<?php 
    echo $view['assets']->getUrl('images/logo.png'); 
    // echoes --->  /assets/images/logo.png
?>

However, my views are much complex and I want to divide different sections of view in functions. So when I write the above code in a function, it doesn't work.

function one(){
    echo $view['assets']->getUrl('images/logo.png'); 
}
one();
Notice: Undefined variable: view in ....\Resources\views\Section\splash.html.php on line 12

Fatal error: Call to a member function getUrl() on a non-object in ....\Resources\views\Section\splash.html.php on line 12

Can someone please guide me how can I get this to work?

This is my full view file.

<?php
echo $view['assets']->getUrl('images/logo.png') . "<br><br>";

function one(){
    echo $view['assets']->getUrl('images/logo.png') . "<br><br>";
}
one();
?>
<img src="<?php echo $view['assets']->getUrl('images/logo.png') ?>" alt="no image" />

This is how I am calling the view from my controller

return $this->render('MySimpleBundle:Section:splash.html.php');

@Nextar Yes I have. My php code <?php echo $view['assets']->getUrl('images/logo.png'); // echoes ---> /assets/images/logo.png ?> works. — Nis
– Nis, Commented Jun 24, 2014 at 5:47

Nextar · Accepted Answer · 2014-06-24 08:11:10Z

1

The function "one()" has his own variable scope, so you have to pass the url into the function http://php.net/manual/en/language.variables.scope.php

function one($url)
{
    echo $view['assets']->getUrl($url) . "<br><br>";
}

anyway, in my opinion you should use simply:

<img src="<?php echo $view['assets']->getUrl('images/logo.png') ?>" alt="Symfony!" />

to render the image, if your view gets to complex, it is may the right time to switch to a powerful templating engine like twig

answered Jun 24, 2014 at 8:11

Nextar

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1 Comment

Nis Over a year ago

Thanks. I don't know how I missed the basic variable scope issue? I had realized it short time after asking but +1 for answering.

pnx · Accepted Answer · 2014-06-24 10:13:25Z

1

Actually, it is the variable $view as the notice says.

You should either pass $view as a parameter or declare it global inside the function scope:

function one($view) {
    echo $view['assets']->getUrl('images/logo.png');
}

or

function one() {
    global $view;
    echo $view['assets']->getUrl('images/logo.png');
}

answered Jun 24, 2014 at 10:13

pnx

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3 Comments

Nextar Over a year ago

This is mostly my answer copied, anyway using global variables is a bad practise and should ever be avoided

Nis Over a year ago

Thanks. I don't know how I missed the basic variable scope issue? I had realized it short time after asking but +1 for answering.

pnx Over a year ago

@Nextar oh yeah, sorry about that. i should have placed a comment instead. And global variables is indeed bad practise and should have been mentioned :)

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