I have this code situation
{% if 1|floatformat in '7,10' %}
yes
{% else %}
no
{% endif %}
the return result is always set to "yes", how to make the result return "no"
please help.
thanks
It's hard to make out from your example what exactly is happening here, because you're substituting dummy data for what's really coming out of the database.
As you've written it, the result will always be "yes" because the string '7,10' contains the string '1'.
It sounds like what you're trying to achieve is:
If this number is in this list, then yes, otherwise no.
So let's rewrite this template to be a little more real:
{% if mynumber in yeslist %} yes {% else %} no {% endif %}
This assumes that:
mynumber is a number
yeslist is a list of numbers
I'm not sure what you're using floatformat for in this case.
If the above assertions aren't true, and you have to use strings, then your work is much harder, and you should be processing yeslist server-side. For example, if yeslist is just a string like "7,10,123,93,9,19,83", then figuring out if the number 8 is in the list will be needlessly difficult in templates. Much easier to do it in your view:
def myview(request):
ctx = {}
# ... do some work ...
# yeslist now has a string like "7,10,123,93,9,19,83"
ctx['yeslist'] = yeslist.split(',')
# ... do more work, and render the response ...
Now, {% if '8' in yeslist %} will no longer return a false positive, because it's not doing a substring match, it's doing list membership.
1|floatformat returns 1 and the condition checks if 1 is in '7,10', since django considers '7,10' as a string it returns True. Try passing a list and it will return no
Related
Seems like elementary question, and yet can't get it work
{% if iterator.next > 10 %}
Do smth
{% endif %}
Two issues. First, this code just won't work (the code in the if-condition never implemented even when the condition seems to hold true), and second issue - the ">" sign is highlighted as if it where the closing tag of the closest open tag. Any ideas how to fix the first issue and is it all right with second issues ? Maybe there's some elegant syntax that I am missing and that would remove this ambiguity for the text editor ?
iterator.next may be a string which would result in the statement being False.
Try creating a custom filter to convert it to an int. For example create "my_filters.py":
# templatetags/my_filters.py
from django import template
register = template.Library()
#register.filter()
def to_int(value):
return int(value)
Then in your template:
{% load my_filters %}
{% if iterator.next|to_int > 10 %}
Do smth
{% endif %}
More on custom tags and filters here
I wouldn't worry about the highlighting, this may just be your IDE. I recommend using PyCharm for Django development
Django's docs says that you can use > with if tag:
{% if somevar < 100 %}
This appears if variable somevar is less than 100.
{% endif %}
take a look at documentation: https://docs.djangoproject.com/en/1.9/ref/templates/builtins/
maybe you are missing something else?
Looks like a question that should be already covered, but after spending some time, I did not find how to check that a variable is numeric in Django template.
Something like
{% if my_var.isnumeric %}
# do something
{% endif %}
UPDATE
As I learnt from the below discussion, there seems to be no built-in tag to check this, and we end up having to create our own template tag.
Assuming that "numeric" means "contains only digits" (and no decimal point, no minus sign, etc.)
Custom filter is your best bet:
from django import template
register = template.Library()
#register.filter(is_safe=True)
def is_numberic(value):
return "{}".format(value).isdigit()
Docs about custom template filters: https://docs.djangoproject.com/en/1.9/howto/custom-template-tags/
Usage in templates:
{% load your_custom_lib %}
...
{% if something|is_numberic %}...
If you consider integers as numeric (positive and negative), then the function becomes:
try:
int("{}".format(value))
except ValueError:
return False
else:
return True
In case "numeric" means "integer or float", then use float instead of int. But note that this will also recognize -12E3 as numeric, because:
>>> -12E3
-12000.0
Does this work ?
{{ value|divisibleby:"1" }}
EDIT: Nope, raises an exception if a string is given.
This seems such a simple thing but I can't seem to get it to work for one, and two I can't seem to be getting a straight forward answer online on wether it can or can't be done.
I just want a simple tag that will work like so
{% if my_tag %}
render something
{% else %}
render something else
{% endif %}
now I don't care about filters or any other ways you might be able to do the same thing, I want it to look exactly like that, and work like described, I have a simple tag made that actually does return True or False as needed, and it's called if I call the tag like this
{% my_tag %}
however it does't get called if I add an if in front of the tag, is a feature this simple and logical not implemented?
Like the link to the potential duplicate states, you can use an assignment tag similar to the following that will return the current time:
#register.assignment_tag
def get_current_time(format_string):
return datetime.datetime.now().strftime(format_string)
Within your template you can then do what you are desiring to do:
{% if get_current_time %}
...show time
{% else %}
...don't show time
{% endif %}
I have a model field, which stores a list of URLs (yeah, I know, that's wrong way) as url1\nurl2\nurl3<...>. I need to split the field into an array in my template, so I created the custom filter:
#register.filter(name='split')
def split(value, arg):
return value.split(arg)
I use it this way:
{% with game.screenshots|split:"\n" as screens %}
{% for screen in screens %}
{{ screen }}<br>
{% endfor %}
{% endwith %}
but as I can see, split doesn't want to work: I get output like url1 url2 url3 (with linebreaks if I look at the source). Why?
Django intentionally leaves out many types of templatetags to discourage you from doing too much processing in the template. (Unfortunately, people usually just add these types of templatetags themselves.)
This is a perfect example of something that should be in your model not your template.
class Game(models.Model):
...
def screenshots_as_list(self):
return self.screenshots.split('\n')
Then, in your template, you just do:
{% for screen in game.screenshots_as_list %}
{{ screen }}<br>
{% endfor %}
Much more clear and much easier to work with.
Functionality already exists with linkebreaksbr:
{{ value|linebreaksbr }}
https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#linebreaksbr
Hm, I have partly solved this problem. I changed my filter to:
#register.filter(name='split')
def split(value, arg):
return value.split('\n')
Why it didn't work with the original code?
I wanted to split a list of words to get a word count, and it turns out there is a filter for that:
{{ value|wordcount }}
https://docs.djangoproject.com/en/dev/ref/templates/builtins/?from=olddocs#wordcount
Apart from whether your original solution was the right approach, I guess the original code did not work because the meaning of the \n is not the same in Python code as it is in HTML: In Python code it means the escaped newline character, in HTML it is just the two separate characters \ and n.
So passing as input parameter \n from the HTML template to the Python code is equivalent to splitting on the Python string \\n: a literal \ followed by a n.
I am trying to pass in url parameters to a django template like this...
response = render_to_string('persistConTemplate.html', request.GET)
This the calling line from my views.py file. persistConTemplate.html is the name of my template and request.GET is the dictionary that contains the url parameters.
In the template I try to use one of the parameters like this...
{% for item in (numItems) %}
item {{item}}
{% endfor %}
numItems is one of the url parameters that I am sending in my request like this...
http:/someDomain/persistentConTest.html/?numItems=12
When I try the for loop above, I get an output like this....
image 1 image 2
I am expecting and would like to see the word image printed 12 times...
image 1 image 2 image 3 image 4 image 5 image 6 image 7 image 8 image 9 image 10 image 11 image 12
Can anyone please tell me what I am going wrong?
you can coerce a str to an int using the add filter
{% for item in numItems|add:"0" %}
https://docs.djangoproject.com/en/dev/ref/templates/builtins/#add
to coerce int to str just use slugify
{{ some_int|slugify }}
EDIT: that said, I agree with the others that normally you should do this in the view - use these tricks only when the alternatives are much worse.
I like making a custom filter:
# templatetags/tag_library.py
from django import template
register = template.Library()
#register.filter()
def to_int(value):
return int(value)
Usage:
{% load tag_library %}
{{ value|to_int }}
It is for cases where this cannot be easily done in view.
Yes, the place for this is in the view.
I feel like the above example won't work -- you can't iterate over an integer.
numItems = request.GET.get('numItems')
if numItems:
numItems = range(1, int(numItems)+1)
return direct_to_template(request, "mytemplate.html", {'numItems': numItems})
{% for item in numItems %}
{{ item }}
{% endfor %}
The easiest way to do this is using inbuilt floatformat filter.
For Integer
{{ value|floatformat:"0" }}
For float value with 2 precision
{{ value|floatformat:"2" }}
It will also round to nearest value. for more details, you can check https://docs.djangoproject.com/en/1.10/ref/templates/builtins/#floatformat.
You should add some code to your view to unpack the GET params and convert them to the values you want. Even if numItems were an integer, the syntax you're showing wouldn't give you the output you want.
Try this:
ctx = dict(request.GET)
ctx['numItems'] = int(ctx['numItems'])
response = render_to_string('persistConTemplate.html', ctx)
In my case one of the items was a string and you can not compare a string to an integer so I had to coerce the string into an integer see below
{% if questions.correct_answer|add:"0" == answers.id %}
<span>Correct</span>
{% endif %}
You can do like that: if "select" tag used.
{% if i.0|stringformat:'s' == request.GET.status %} selected {% endif %}
My solution is kind of a hack and very specific..
In the template I want to compare a percentage with 0.9, and it never reaches 1, but all the values are considered string in the template, and no way to convert string to float.
So I did this:
{% if "0.9" in value %}
...
{% else %}
...
{% endif %}
If I want to detect some value is beyond 0.8, I must do:
{% if ("0.9" in value) or ("0.8" in value) %}
...
{% else %}
...
{% endif %}
This is a hack, but suffice in my case. I hope it could help others.