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 provide me with the information.

<li {% if page=="home" %}class="active"{% endif %}>                   
    <a href="/">Home</a>                                                
</li>                                                                 
<li {% if page=="about" %}class="active"{% endif %}>                  
    <a href="/about">About</a>                                          
</li>

score 73 · Accepted Answer · answered Dec 30 '11 at 05:37

73

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

answered Dec 30 '11 at 05:37

ranksrejoined

1,229
9
9

6

Note that this is not true for other variables. It works because `request` is one of the variables [inserted by default](http://flask.readthedocs.org/en/latest/templating/#standard-context) into the template context. – Marcel M Mar 25 '13 at 14:39

score 64 · Answer 2 · answered Jul 29 '14 at 10:27

64

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!

answered Jul 29 '14 at 10:27

negus

801
6
7

7

This is the correct answer. The others use hardcoded paths, which is a very bad idea. – Nathan Osman Dec 09 '14 at 19:53
2

This only works you are using one route per section of the page. Otherwise this won't be what you are looking for. =\ – flamusdiu May 28 '15 at 02:22
1

No, that's not true. You can have some structure in your endpoint names (like blueprint.subsection.page) and make the comparison using .startswith('blueprint.subsection.'). Just an example. – negus May 29 '15 at 03:25

score 50 · Answer 3 · edited Sep 23 '15 at 17:45

50

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>

edited Sep 23 '15 at 17:45

Community

1
1

answered Jan 03 '12 at 16:59

Sosiska

609
4
2

2

`request` imported (into `*.py` file), still getting error: jinja2.exceptions.UndefinedError: 'request' is undefined – Pavel Vlasov Feb 09 '18 at 21:15

score 11 · Answer 4 · answered Mar 14 '16 at 11:18

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():

score 5 · Answer 5 · edited Jul 16 '15 at 11:14

5

Try

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

This one worked for me.

edited Jul 16 '15 at 11:14

bummi

27,123
14
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101

answered Jul 16 '15 at 10:44

iChux

2,266
22
37

score 1 · Answer 6 · answered Sep 01 '19 at 08:57

1

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"

answered Sep 01 '19 at 08:57

Iaron

11
1

Flask: current page in request variable

6 Answers6

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