Flask: current page in request variable

Question

In a template, how do I get what page I\'m currently on? I\'d rather not pass a variable like page , especially when I know some request.xxx can pr

Answer 1

Try

<li {% if request.endpoint == "blueprintname.routename" %}class="active"{% endif %}>Home</li>

This one worked for me.

Answer 2

To avoid using hard-coded URLs you can use the url_for function like this:

{% for ni in ['index', 'foo', 'bar', 'baz'] %}
<li {%- if request.path == url_for(ni) %} class="active"{% endif %}><a href="{{ url_for(ni) }}">{{ ni | capitalize }}</a></li>
{% endfor %}

In this case index, foo, bar and baz would be function names, used like this in your python code:

@app.route('/')
def index():

Answer 3

Using request.path doesn't seem to be a proper approach since you'll have to update the paths in case of changing URL rules or deploying your site under a subfolder.

Use request.url_rule.endpoint instead, it contains actual endpoint name independent of actual path:

(Pdb) request.url_rule.endpoint
'myblueprint.client_pipeline'

In a template:

<li {% if request.url_rule.endpoint == "myblueprint.client_pipeline" %}class="active"{% endif %}>Home</li>

Good luck!

Answer 4

As long as you've imported request, request.path should contain this information.

Answer 5

First import request from flask in your application. Then you can use it without passing to template:

<li {%- if request.path == "/home" %} class="active"{% endif %}>
    <a href="/">Home</a>
</li>
<li {%- if request.path=="/about" %} class="active"{% endif %}>
    <a href="/about">About</a>
</li>

Answer 6

You can also use .split if your url has more stuff in it.

Example:

/product
/product/add
/product/32432/edit
/product/32432/view

{{ request.path.split('/')[1] }}

This will return only "product"