The logic for showing a tag
{% if profile.facebook is not None %}<i class="facebook icon"></i>{% endif %}
i want to show the facebook icon only when the user has provided the link. if not i dont want to display the icon.
You can define your conditional as well:
{% if profile.facebook %}
<i class="facebook icon"></i>
{% endif %}
if your code not work, you can share more info, like your views.py etc.
Related
How do I create a template filter that only shows a section of code on a specified page? For example, my homepage which is my base_generic has code that is specific to that page. Whenever another page is loaded that code should not be displayed.
I know I can use the block system but was wondering if I could also accomplish the task with a combination of the URL and If filter. See code below:
# Code for base_generic template
{% url 'organizer-homepage' as home_page %}
{% if home_page %}
Shows Content Only on Specified Page
{% else %} Show Alternate on All Other Pages
{% endif %}
You can do something like:
{% url 'organizer-homepage' as home_page %}
{% if request.path == home_page %}
Shows Content Only on Specified Page
{% else %} Show Alternate on All Other Pages
{% endif %}
Also make sure you have django.core.context_processors.request in your template context processors, which is there by default.
I am pretty new to Flask/Flask-Admin.
I have followed the tutorial on flask admin and managed to get the admin panel working but slightly lost on how to get the below things implemented.
https://github.com/flask-admin/flask-admin/tree/master/examples/auth
When logged in as a normal user I can only see "home" page.
How can I expose other views to "normal user" and restrict actions such as read only etc.
I have created a "baseview" which is not associated with any other models as below:
class SitesView(MyBaseView):
#expose('/')
def index(self):
return self.render('views/testviews.html')
admin.add_view(SitesView(name='Test views', endpoint='test views'))
and html as below:
{% extends 'admin/master.html' %}
{% block body %}
{{ super() }}
{% if current_user.has_role('view1') %}
Site1
{% endif %}
{% if current_user.has_role('view2') %}
<a>Site2</a>
{% endif %}
{% if current_user.has_role('view3') %}
<a>Site3</a>
{% endif %}
{% if current_user.has_role('view4') %}
<a>Site4</a>
{% endif %}
{% endblock %}
This gives me a new tab with different views with works as expected.
What I am trying to achieve here is when user click the Site1 link they go to Site1 page within flask-admin interface but I am not sure how to do that. I could create a new route for this but the problem is I can't(don't know how to) extend flask admin template.
For example this works but it redirect the page outside flask-admin template:
#app.route('/views/')
def views():
return render_template('views/views1.html')
and modified the templates>admin>index.html page with below:
<ul class="lead text-center list-group">
{% if current_user.has_role('view1') %}
<li class="list-group-item">View1</li>
{% endif %}
{% if current_user.has_role('view2') %}
<li class="list-group-item">View2</li>
{% endif %}
{% if current_user.has_role('view3') %}
<li class="list-group-item">View3</li>
{% endif %}
{% if current_user.has_role('view4') %}
<li class="list-group-item">View4</li>
{% endif %}
</ul
I want to build the whole web site using flask admin so that I can keep user experience consistence. Am I doing this the wrong way?
Thanks for your time.
Please do let me know if you want me to provide more information on this issue.
Kind Regards.
So after going through documentations and tutorials I have found the solution to my issue.
For my first question:
When logged in as a normal user I can only see "home" page. How can I
expose other views to "normal user" and restrict actions such as read
only etc.
We can do this by overwriting our view functions is_accessible method as below:
def is_accessible(self):
if not current_user.is_active or not current_user.is_authenticated:
return False
if current_user.has_role('superuser') or current_user.has_role('user') or current_user.has_role('view1'):
return True
return False
For my second question we just need to give the endpoint as for our BaseView as below:
class MyView(BaseView):
#expose('/')
def index(self):
return self.render('views.html')
admin.add_view(MyView(name='Custom Views', endpoint='customviews'))
And then in your jinja template you need to call it:
href="{{ url_for('customviews.index') }}
Just one thing to note, doing this:
current_user.has_role('superuser') or current_user.has_role('user') or current_user.has_role('view1')
could get quite messy if we have so many roles, not sure how we would approach this but hoping this will help someone.
Thanks all.
I know this is an old question, but for the following code
current_user.has_role('superuser') or current_user.has_role('user') or current_user.has_role('view1')
What I like to do is having a hybrid_property (available on both Peewee and SQLAlchemy) inside my User class that consolidates these properties. So it'd look something like this:
#hybrid_property
def user_has_administrative_rights(self):
return self.has_role('superuser') or self.has_role('user')
I would like to create a product that will be available in Shopify's storefront but only accessible for the shop administrator. Is there a way to identify if the current user is an admin via liquid? Or is there any other solution for this. Thanks!
If you're signed in as an admin, when rendering the {{ content_for_header }} include, it will contain some JavaScript to push the page content down to make room for the Shopify admin bar.
We can utilize capture to store the {{ content_for_header }} code as a liquid variable and then use the contains operator to check if admin_bar_iframe exists in the variable.
{% capture CFH %}{{ content_for_header }}{% endcapture %}{{ CFH }}
{% if CFH contains 'admin_bar_iframe' %}
{% assign admin = true %}
{% endif %}
{% if admin %}
<!-- User is an admin -->
{% else %}
<!-- User is not an admin -->
{% endif %}
Note: I've noticed that the Shopify admin bar doesn't populate at all times (I think its a bug). If your Shopify admin bar is not populating on your instance this will not work.
Just figured out this method that works. (Basically the same thing as the old method, just another way around!)
Detecting logged in admin viewing site:
{% if content_for_header contains 'adminBarInjector' %}
<script>
console.log("You're a logged in admin viewing the site!");
</script>
{% endif %}
Detecting admin in design mode:
{% if content_for_header contains 'designMode' %}
<script>
console.log("You're an admin in design mode!");
</script>
{% endif %}
Another approach would be to use Customer Accounts. Liquid provides a {{ customer }} object, which is only present when a user (customer) is logged in.
You can add a specific tag to an admin user and use liquid to verify if a tag is present:
{% if customer.tags contains "admin" %}
And of course you need to identify your product as 'admin-only', for example using tags:
{% if customer.tags contains "admin" and product.tags contains "admin" %}
<!-- render product -->
{% else %}
<!-- do nothing -->
{% endif %}
EDIT: This seems not to be working anymore since an update to Shopify.
I know this is late, but here is what I've used and it has worked correctly in my testing. This is adapted from the previous answers, but allows use for Customise Theme options, etc.
{% capture CFH %}{{ content_for_header }}{% endcapture %}{{ CFH }}
{% assign isAdmin = true %}
{% if CFH contains '"__st"' %}
{% if CFH contains 'admin_bar_iframe' %}{% else %}
{% assign isAdmin = false %}
{% endif %}
{% endif %}
This is a more complete version of that provided by kyle.stearns above (I can't comment on it because new). One of the main instances where admin bar doesn't load is when previewing themes. Here is the amended code which I've used (updated June 2018 - as Shopify edited the way we preview themes):
{% capture CFH %}{{ content_for_header }}{% endcapture %}
{% if CFH contains 'admin_bar_iframe' %}
{% assign admin = true %}
{% elsif CFH contains 'preview_bar_injector-' %}
{% assign admin = true %}
{% endif %}
{% if admin %}
<!-- User is an admin -->
<script>
alert ("do some work");
</script>
{% else %}
<!-- User is not an admin -->
<script>
alert ("please buy some stuff");
</script>
{% endif %}
If you're using Plus then you also have access to the User via api.
I know this is an old question, but it still shows on top in google searches.
There's now a much better way.
Using liquid:
{% if request.design_mode %}
<!-- This will only render in the theme editor -->
{% endif %}
Using javascript:
if (Shopify.designMode) {
// This will only render in the theme editor
}
Source: shopify.dev
Currently there isn't. You could perhaps try inspecting cookies and stuff to see if there's some identifying information that would let you know if the user is an admin, but it would be fragile.
This would also require rendering the items, but hiding them via CSS. Then you'd show them using JS after you've run your checks.
As stated, this would probably be really fragile.
I'm building a website using django with a header on top of every page, which basically is a menu with a few links, constant throughout the pages.
However, depending on the page you're on I'd like to highlight the corresponding link on the menu by adding the class "active". To do so, I am currently doing as follow: each page has a full menu block that integrates within a general layout, which does NOT contain the menu. For exemple, page2 would look like this:
{% extends "layout.html" %}
{% block menu %}
<li>Home</li>
<li>page1</li>
<li class="active">page2</li>
<li>page3</li>
{% endblock %}
The problem is that, beside from that solution being not so pretty, every time I want to add a link to the header menu I have to modify each and every page I have. Since this is far from optimal, I was wondering if any of you would know about a better way of doing so.
Thanks in advance!
You can create a custom templatetag:
from django import template
from django.core.urlresolvers import reverse, NoReverseMatch, resolve
register = template.Library()
#register.simple_tag
def active(request, view_name):
url = resolve(request.path)
if url.view_name == view_name:
return 'active'
try:
uri = reverse(view_name)
except NoReverseMatch:
uri = view_name
if request.path.startswith(uri):
return 'active'
return ''
And use it in the template to recognize which page is loaded by URL
<li class="{% active request 'car_edit' %}">Edit</li>
If you have a "page" object at every view, you could compare a navigation item's slug to the object's slug
navigation.html
<ul>
{% for page in navigation %}
<li{% ifequal object.slug page.slug %} class="active"{% endifequal %}>
{{ page.title }}
</li>
{% endfor %}
</ul>
base.html
<html>
<head />
<body>
{% include "navigation.html" %}
{% block content %}
Welcome Earthling.
{% endblock %}
</body>
</html>
page.html
{% extends "base.html" %}
{% block content %}
{{ object }}
{% endblock %}
Where navigation is perhaps a context_processor variable holding all the pages, and object is the current PageDetailView object variable
Disclaimer
There are many solutions for your problem as noted by Paulo. Of course this solution assumes that every view holds a page object, a concept usually implemented by a CMS. If you have views that do not derive from the Page app you would have to inject page pretenders within the navigation (atleast holding a get_absolute_url and title attribute).
This might be a very nice learning experience, but you'll probably save loads time installing feinCMS or django-cms which both define an ApplicationContent principle also.
You may use the include tag and pass it a value which is the current page.
For example, this may be a separate file for declaring the menu template only:
menu.html
{% if active = "a" %}
<li>Home</li>
{% if active = "b" %}
<li>page1</li>
{% if active = "c" %}
<li class="active">page2</li>
{% if active = "d" %}
<li>page3</li>
And call this from within your template like this:
{% include 'path/to/menu.html' with active="b"%} # or a or c or d.
Hope it helps!
I'm working on a site where the footer content is shared across all pages. What is the best way to do in Django-CMS?
I tried using show_placeholder tag, but it somehow didn't work. A little more details on what I did:
First, I have a {% placeholder footer_info %} in base.html. Then I add a page called "Home" (template homepage.html) in django admin and put some text under footer_info as a Text plugin. As the accepted answer in this question suggested (http://stackoverflow.com/questions/3616745/how-to-render-django-cms-plugin-in-every-page),
I add
{% placeholder footer_info or %}
{% show_placeholder footer_info "Home" %}
{% endplaceholder %}
In a template called services.html which I used as the template for page Services.
However, the content in home page is not showing up in services page. I also tried adding an id home_cms_page to home page in the Advanced option area, so that I can reference it in services.html like this:
{% placeholder footer_info or %}
{% show_placeholder footer_info "home_cms_page" %}
{% endplaceholder %}
But the content is still not showing up.
Could anyone tell me what I am doing wrong? And is this the best way of getting some content from a page across all other pages (and I have to add show_placeholder in every other page)?
Thank you
EDIT:
It is not a multilingual site. I commented out 'cms.middleware.multilingual.MultilingualURLMiddleware', because the only language I use on the site is English.
I have this in my base.html:
{% load cms_tags sekizai_tags %}
<!-- all the rest of the HTML markups -->
<div class="span4">
{% placeholder footer_info %}
</div>
Then I added a page in the admin called "Home" with a Text plugin and an id of "home_cms_page".
The following is in my services.html:
{% extends "base.html" %}
{% load cms_tags %}
{% block base_content %}
{% placeholder services_info %}
{% endblock base_content %}
{% block page_content %}
Home page
{% endblock page_content %}
{% placeholder "footer_info" or %}
{% show_placeholder "footer_info" "home_cms_page" %}
{% endplaceholder %}
Read the documentation:
If you know the exact page you are referring to, it is a good idea to
use a reverse_id (a string used to uniquely name a page) rather than a
hard-coded numeric ID in your template. For example, you might have a
help page that you want to link to or display parts of on all pages.
To do this, you would first open the help page in the admin interface
and enter an ID (such as help) under the ‘Advanced’ tab of the form.
Then you could use that reverse_id with the appropriate templatetags:
{% show_placeholder "right-column" "help" %}
I added "index" in the advanced options of the index page, and added {% show_placeholder "banner" "index" %} in the base template. It all works.