'Check if variable is string or array in Twig
Is it possible to check if given variable is string in Twig ?
Expected solution:
messages.en.yml:
hello:
stranger: Hello stranger !
known: Hello %name% !
Twig template:
{% set title='hello.stranger' %}
{% set title=['hello.known',{'%name%' : 'hsz'}] %}
{% if title is string %}
{{ title|trans }}
{% else %}
{{ title[0]|trans(title[1]) }}
{% endif %}
Is it possible to do it this way ? Or maybe you have better solution ?
Solution 1:[1]
Ok, I did it with:
{% if title[0] is not defined %}
{{ title|trans }}
{% else %}
{{ title[0]|trans(title[1]) }}
{% endif %}
Ugly, but works.
Solution 2:[2]
I found iterable to not be good enough since other objects can also be iterable, and are clearly different than an array.
Therefore adding a new Twig_SimpleTest to check if an item is_array is much more explicit. You can add this to your app configuration / after twig is bootstrapped.
$isArray= new Twig_SimpleTest('array', function ($value) {
return is_array($value);
});
$twig->addTest($isArray);
Usage becomes very clean:
{% if value is array %}
<!-- handle array -->
{% else %}
<!-- handle non-array -->
{% endif % }
Solution 3:[3]
There is no way to check it correctly using code from the box.
It's better to create custom TwigExtension and add custom check (or use code from OptionResolver).
So, as the result, for Twig 3, it will be smth like this
class CoreExtension extends AbstractExtension
{
public function getTests(): array
{
return [
new TwigTest('instanceof', [$this, 'instanceof']),
];
}
public function instanceof($value, string $type): bool
{
return ('null' === $type && null === $value)
|| (\function_exists($func = 'is_'.$type) && $func($value))
|| $value instanceof $type;
}
}
Solution 4:[4]
Assuming you know for a fact that a value is always either a string or an array:
{% if value is iterable and value is not string %}
...
{% else %}
...
{% endif %}
This worked good enough for me in a project I was working on. I realize you may need another solution.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | hsz |
| Solution 2 | |
| Solution 3 | wtorsi |
| Solution 4 | Darren Embry |
