I have a Django template that looks something like this:
{% if thing in ['foo', 'bar'] %}
Some HTML here
{% else %}
Some other HTML
{% endif %}
The problem is it comes back empty. If I switch to this:
{% if thing == 'foo' or thing == 'bar' %}
Some HTML here
{% else %}
Some other HTML
{% endif %}
it works fine. Is there some reason you can't use x in list in Django templates?
You can. But you can't use a list literal in templates. Either generate the list in the view, or avoid using if ... in ....
I got it working with the help of this answer. We could use split to generate a list inside the template itself. My final code is as follows (I want to exclude both "user" and "id")
{% with 'user id' as list %}
{% for n, f, v in contract|get_fields %}
{% if n not in list.split %}
<tr>
<td>{{f}}</td>
<td>{{v}}</td>
</tr>
{% endif %}
{% endfor %}
{% endwith %}
Send the list from the context data in the view.
Views.py:
class MyAwesomeView(View):
...
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
context['list'] = ('foo', 'bar')
...
return context
MyTemplate.html:
{% if thing in list %}
Some HTML here
{% else %}
Some other HTML
{% endif %}
Tested on Django version 3.2.3.
There is a possibility to achieve that also by creating a custom filter:
Python function in your_tags.py:
from django import template
register = template.Library()
#register.filter(name='is_in_list')
def is_in_list(value, given_list):
return True if value in given_list else False
and passing your list to html django template:
{% load your_tags %}
{% if thing|is_in_list:your_list %}
Some HTML here
{% else %}
Some other HTML
{% endif %}
or without passing any list - by creating a string containing your list's values (with a filter you still can't use list literal), e.g.:
{% load your_tags %}
{% if thing|is_in_list:'foo, bar' %}
Some HTML here
{% else %}
Some other HTML
{% endif %}
[TIP] Remember not to insert space after :.
Related
Hi2all!
I have:
1) mongo collection:
[{_id:ObjectId("5b0d5fb624d22e1b4843c06b")
collectionName:"collection0"
collectionCaption:"caption1"}
{_id:ObjectId("5b0d5fb824d22e1b4843d4c1")
collectionName:"collection1"
collectionCaption:"caption1"}
{_id:ObjectId("5b0d5fb924d22e1b4843d74a")
collectionName:"collection2"
collectionCaption:"caption1"}
{_id:ObjectId("5b0d5fb924d22e1b4843d7b0")
collectionName:"collection3"
collectionCaption:"caption1"}]
2) flask app with the view:
def index():
a = mongo.db.collectionsNames.find()
return render_template('index.html', collectionsNames=a)
3) templates: index.html witch extend base.html.
base.html:
{% extends "bootstrap/base.html" %}
{% block content %}
<div class="dropdown-menu" aria-labelledby="navbarDropdownMenuLink">
{% for asd in collectionsNames %}
<a class="dropdown-item" href="/{{ asd["collectionName"] }}">{{ asd["collectionCaption"] }}</a>
{% endfor %}
</div>
{% block contentBase %} {% endblock %}
{% endblock %}
index.html:
{% extends "base.html" %}
{% block contentBase %}
{% for zxc in collectionsNames %}
{{ zxc["collectionName"] }}
{% endfor %}
{% endblock %}
The question is: why if base and index using the same variable of collection, index template show nothing?
But if in view code is:
def index():
a = mongo.db.collectionsNames.find()
b = mongo.db.collectionsNames.find()
return render_template('index.html', collectionsNames1=a, collectionsNames2=b)
and in templates i am using the different variables, the index template show me the data.
mongo.db.collectionsNames.find() returns a cursor: print(mongo.db.collectionsNames.find()) gives <pymongo.cursor.Cursor object at 0x7fd3854d5710>.
To make it easy (but kind of wrong) a cursor is a specific kind of instance that fetch data from the database chunk by chunk so if you want the 1 000 000 first items in db you don't actually store 1M items in RAM, you iterate 100 by 100 items (for example). Cursor handle that in a magic way.
Whatever, you cannot loop several time on the same cursor and you should never caste a cursor to list like list(cursor_instance) (because if your query ask for 1M product, doing this add all these products in RAM).
So, now, how can you handle this. Most of time I would say it's better to call the method when you need it, twice if needed.
But here you are in a Jinja environment and if I'm right calling methods within a Jinja template is impossible.
A way to do that is to use properties.
class LazyCollection(object):
#property
def collections_names(self):
return mongo.db.collectionsNames.find()
lazycoll = LazyCollection()
return render_template('index.html', collectionsNames=lazycoll)
Then, in your template:
{% for asd in collectionsNames.collections_names %}
<p>{{ asd.foo }}</p>
{% endfor %}
I have a context dictionary entry objectives that maps objective query objects to a list of tests that belong to that objective. Example code:
objectives = Objective.objects.filter(requirement=requirement)
context_dict["requirements"][requirement] = objectives
for objective in objectives:
tests = Test.objects.filter(objective=objective)
context_dict["objectives"][objective] = tests
In my django html template, I iterate over objectives and display them. I then want to iterate over the tests that belong to these objectives. When I do this:
{% for test in {{ objectives|get_item:objective }} %}
I get a TemplateSyntaxError: 'for' statements should use the format 'for x in y':
In the application/templatetags directory, I have:
from django.template.defaulttags import register
...
#register.filter
def get_item(dictionary, key):
return dictionary.get(key)
If instead I make {{ objectives|get_item:objective }} a JS variable, I see that it does indeed produce a list, which I should be able to iterate over. Of course, I can't mix JS variables and the django template tags, so this is only for debugging:
var tests = {{ objectives|get_item:objective }}
var tests = [<Test: AT399_8_1>, <Test: AT399_8_2>, <Test: AT399_8_3>, <Test: AT399_8_4>, <Test: AT399_8_5> '...(remaining elements truncated)...']
How do I iterate over this list in the django template tag?
You cannot user {{...}} inside the {%...%}
What you can try is changing your filter to an assignment tag and using that value in the loop
#register.assignment_tag
def get_item(dictionary, key):
return dictionary.get(key)
And then in your template use it as
{% get_item objectives objective as tests %}
{% for test in test %}
....
{% endfor %}
Instead of all this if your models are proper with foreign keys I would do something like
{% for objective in requirement.objective_set.all %}
{% for test in objective.test_set.all %}
....
{% endfor %}
{% endfor %}
In my context I would pass only the requirement
You already have an answer, but note that dropping the {{ }} tags and keeping everything else the same would have worked fine.
{% for test in objectives|get_item:objective %}
**This is Right Answer for Using Django if else and for loop **
Note :- We Have to Put Key in " " string (Double quotes) some time produce an error so That is good way bcz i faced that problem whwn i Learned
{% if 'numbers'|length > 0 %}
{% for i in numbers %}
{% if i > 20 %}
{{i}}
{% endif %}
{% endfor %}
{% else %}
Empty
{% endif %}
I am working on a small web app to view some logfiles. But the
queries I issue to the database use to get very big.
I wanted to implement some pagination following this example pagination.
I put the class into a single file which gets loaded in the Flask view. Next I implemented
my pagination view like this:
#app.route('/index/', defaults={'page':1})
#app.route('/index/page/<int:page>')
def index(page):
count = db_session.execute("select host,facility,level,msg from messages").rowcount
tblqry = db_session.execute("select host,facility,level,msg from messages").fetchmany(size=1000)
if not tblqry and page != 1:
abort(404)
pagination = Pagination(page, PER_PAGE, count)
return render_template('index.html', pagination=pagination, tblqry=tblqry)
after that I created a macro file named _pagination_helper.html with the macro contents from the macro example. Then I imported the pagination_helper macro with:
{% from "_pagination_helper.html" import render_pagination %}
but when I then try to do something like this:
{{ render_pagination(host[0]) }}
flask claims:
UndefinedError: 'str object' has no attribute 'iter_pages'
so why does flask fails to find the 'iter_pages' because I included the pagination class in the views file?
And also I am not really sure where to put the URL Generation Helper from the How To.
Edit:
This is what my pagination_helper looks like:
{% macro render_pagination(pagination) %}
<div class=pagination>
{% for page in pagination.iter_pages() %}
{% if page %}
{% if page != pagination.page %}
{{ page }}
{% else %}
<strong>{{ page }}</strong>
{% endif %}
{% else %}
<span class=ellipsis>…</span>
{% endif %}
{%- endfor %}
{% if pagination.has_next %}
Next »
{% endif %}
</div>
{% endmacro %}
You are expected to pass a Pagination object to the macro, not a string. host[0] is a string, not the pagination value you created in your view function.
Use:
{{ render_pagination(pagination) }}
I have the following in my html file:
{% trans "Result: "%} {{result}}
Which will print out the word SUCCESS on the browser (because thats what the string contains)
But If I do the following:
{% if result == 'SUCCESS' %}
do something
{% else %}
do something else
{% endif %}
I find that the if statement does not work as expected.
Why is this??
The if statement works fine. Your problem must be regarding the string. Maybe it's not a string at all.
Try the ifequal templatetag:
{% ifequal result 'SUCCESS' %}
do something
{% else %}
do something else
{% endifequal %}
You can try different things. If you're assigning result in a view, you can validate it's a string in that very same view:
def my_view(request):
# ... processing ...
result = something()
# Let's make sure it's a string containing 'SUCCESS'
assert type(result) == str
assert result == 'SUCCESS'
You can apply the same logic if it's a context processor.
https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#ifequal
Check this link:
Django String format.
according to django documentation you should use this format:
{% if result|stringformat:"s" == 'SUCCESS' %}
do something
{% else %}
do something else
{% endif %}
or
{% if result|stringformat:"s" in 'SUCCESS' %}
do something
{% else %}
do something else
{% endif %}
or
{% ifequal result|stringformat:"s" 'SUCCESS' %}
do something
{% else %}
do something else
{% endif %}
this problem happen because of your variable type, you should change it to string before compare it to another string.
I've made a django template tag that counts one of my custom user many-to-many field length:
from django import template
register = template.Library()
#register.simple_tag(takes_context=True)
def unread_messages_count(context):
user = context['request'].user
return len(user.messages_unread.all())
and within the template itself, I want to show it to user only if it's larger than zero, so I tried:
{% ifnotequal unread_messages_count 0 %}
some code...
{% endifnotequal %}
but obviously it didn't work. not even with a 'with' statement:
{% with unread_messages_count as unread_count %}
{% ifnotequal unread_count 0 %}
some code...
{% endifnotequal %}
{% endwith %}
How can I check that the variable is larger than 0 and only if it is, present some code to the user (including the number in the variable itself).
thanks.
The easiest way would be to use an assignment tag..
https://docs.djangoproject.com/en/dev/howto/custom-template-tags/#assignment-tags
#register.assignment_tag(takes_context=True)
def unread_messages_count(context):
user = context['request'].user
return len(user.messages_unread.all())
{% unread_messages_count as cnt %}
{% if cnt %}
foo
{% endif %}
you can use a django custom filter https://docs.djangoproject.com/en/dev/howto/custom-template-tags/#writing-custom-template-filters
def unread_messages_count(user_id):
# x = unread_count ## you have the user_id
return x
and in the template
{% if request.user.id|unread_messages_count > 0 %}
some code...
{% endif %}