My project grows every day, so I decide to put repeating content in a reusable snippet and call the snippet in place:
<?php if($image = $data->image()): ?>
<img src="<?php echo thumb($image, array(
'width' => 480,
'quality' => 1
))->url(); ?>"
data-src="<?php echo $image->url() ?>"
alt="<?php echo $site->title()->html(); ?> – <?php echo html($data->image()->caption()) ?>"
class="lazyload blur-up" />
<?php endif ?>
than I call the snippet in a another snippet:
<?php snippet('image') ?>
But I got an error: Undefined variable: data
How do I send variables with a snippet?
There is an section here: https://getkirby.com/docs/templates/snippets but I did not get it to work 
snippet('nameofsnippet', $variable, false/true);
the last parameter will “echo” or render the content and print it like it is, such as for in emails…
@carstengrimm
<?php snippet('image', $data; true) ?>
Not sure what is wrong, but this gives the following error:
syntax error, unexpected ';', expecting ',' or ')'
you use $data; instead $data,
(simicolon is wrong)
<?php snippet('image', $data) ?>
<?php snippet('image', $data) ?>
did you define your variable $data somewhere? you can’t call a variable if it’s not available
Then you probably have to use the long version:
<?php snippet('image', array('data' => $data)) ?>
provided that the variable used in the snippet, that references the snippet, is called $data.
1 Like
@carstengrimm @texnixe
The long form did the trick!
It is possible to hand over more than one $variable?
BTW: since you have already defined $image, you can then use this variable throughout in your snippet:
<?php echo html($image->caption()) ?>
1 Like
Yes,
<?php snippet('image', array('data' => $data, 'someothervariable' => $someothervariable)) ?>
Edit: you can also use another variable name in the snippet, for example:
<?php snippet('image', array('something' => $someother)) ?>
2 Likes
