如何在Symfony2的树枝中调用函数？

Question

In my model (Task) I have a function:

public function isTaskOverdue()
    {
        if ("now"|date('Y-m-d') > task.deadline|date('Y-m-d')){
            return false;

        } else{
            return true;
        }
    }

In twig (edit) I want to display form:

{% extends 'base.html.twig' %}

{% block title %}app:Resources:Task:edit{% endblock %}

{% block body %}

        {{ form(form) }}

{% endblock %}

I want to display form, if this function return true. How can I call this function in twig?

Answer 1

Pass the task entity to twig and call method from object task :

{% if task.isTaskOverdue %}
    {{ form(form) }}
{% endif %}

Answer 2

I think it should be your controller that receives the function result and display the form or not depending on it.

Also you can write your function like so :

public function isTaskOverdue()
    {
        return ("now"|date('Y-m-d') > task.deadline|date('Y-m-d'));
    }

Answer 3

Pass the task entity to twig and do :

{% extends 'base.html.twig' %}

{% block title %}app:Resources:Task:edit{% endblock %}

{% block body %}

    {% if "now"|date("Ymd") <= task.deadline|date("Ymd") %}

        {{ form(form) }}

     {% endif %}

{% endblock %}

But, caution :

If you just not display the form, there is a security issue, because if an attacker submit the form from an self rebuilded HTML page, your controller will receive the form data and apply it.

So I would do the check in the controller, and only create and pass the form to the twig template if the condition is true.

Then, in twig you can use :

{% if form is defined %}
    {{ form(form) }}
{% endif %}