How do I render specific child entries of a structure based on url segment For simplicity, I'm building a site for a restaurant. The "Menu" page has the following url(s):

mysite.com/food/appetizers
mysite.com/food/sandwiches


food is a structure. It contains every entree the restaurant has. I've nested the actual entree to under the corresponding category. For example:

mysite.com/food/appetizers
                         /cheese-curds
                         /french-fries 

mysite.com/food/sandwiches
                        /whole-sandwich
                        /half-sandwich


..and so on.

On mysite.com/food/ I can render all entrees under thier corresponding category like this:

{% set entries = craft.entries.section('food') %}

 {% for entry in entries %}
       {% if entry.type == 'category' %}  
           &lt;h3&gt;&lt;a href="{{ entry.getUrl }}"&gt;{{ entry.title }}&lt;/a&gt;&lt;/h3&gt;
       {% endif %}

       {% set children = entry.getChildren() %}
       {% for child in children %}
           &lt;h5&gt;&lt;a href="{{ child.getUrl }}"&gt;{{ child.title }}&lt;/a&gt;&lt;/h5&gt;
       {% endfor %}
 {% endfor %}


Everything is working great. All entrees are listed below their corresponding "category" on my /food/ page. I can click on each individual entree "cheese-curds" for example, and can get the entry details.

For templates I have _entry and index. 

index is used on mysite.com/food _entry is used on each entree. 

Where I'm stuck is how to render the entries for the current category. For example:

mysite.com/food/appetizers


I'm not sure How to get past my initial command of getting ALL entires:

 {% set entries = craft.entries.section('food') %}

 {% for entry in entries %}
       {% if entry.type == 'category' %}
       ....
       .... 


To only this:

{% if entry.type == 'category' %}
  ....
  ....


So that I am only listing out the entries for the current category. It's OK if I need to create a new template, I'm not sure how to tell Craft how to do that. Or is there a better approach to getting all entries for the structure? It seems simple enough in theory, I keep talking myself in circles.

EDIT

Thank you @j00lz
Your solution makes perfect sense. I think I either don't quite understand, or I'm still at the same spot I was before.

{% switch entry.level %}
    {% case 1 %}        
        {# this is mysite.com/food #}
        {# outputs #}
        Appetizers
            Cheese Curds
            French Fries
        Sandwiches
            Sandwich One
            Sandwich Two
    {% case 2 %}
        {# this is mysite.com/food/appetizers
        {# outputs #}
        Appetizers
            Cheese Curds
            French Fries
    {% case 3 %}
        {# this is mysite.com/food/appetizers/french-fries
        {# outputs #}
            French Fries
            French fries are awesome!
{% endswitch %}


So, it's like I need another template to only show the 2nd case on the page.
I can list all foods, and the actual entrée (3rd case) using 2 templates. I'm hung up on that middle page where it would only list the title of the entrée, and then the children entrée.

I thought about putting an if statement at the top of the page that checks the URL level. If it's only /food/ do "x", if its /food/appetizers do "y", but that seems to assume an awful lot. 

Maybe there is a better route to go all together. I only had a structure in mind because of the nested elements that I wanted to be able to control where they were.
Probably the simplest way is to split the code up like this in your template.

{% switch entry.level %}
    {% case 1 %}        
        {# all the food #}
    {% case 2 %}
        {# course level #}
    {% case 3 %}
        {# one dish #}
{% endswitch %}


This give you a lot of fine control over each level of your structure.

Or even simpler:

{% set entries = craft.entries.section('food').descendantOf( entry ) %}


and run the same loop regardless of which level you are on.

You can also have a look at this page to see how to create a navigation for a structured section.

And if you want elaborately styled menus check out this link