My layout template contains a header and I'd like to add a CSS class to the currently selected menu option.
class="active"
I thought I found the solution here but I'm still having issues. When I paste the following code into my template I get a server error.
{% rule = request.url_rule %}
Why didn't this work? How can I set a variable to control the menu in the templates?
{% rule = request.url_rule %} is not valid syntax in Jinja. If you want to set a context variable from a template, use set:
{% set rule = request.rule %}
request is already passed to each template context automatically, you don't need to pass it in render_template.
My best guess would be that the request object isn't getting passed to the template. If it isn't being passed you wouldn't be able to access it when rendering the template. You could pass it explicitly, which would look something like this:
from flask import request
from flask import render_template
#app.route('/hello/')
def hello():
return render_template('hello.html', request=request)
Someone commented that swankswashbucklers answer wouldn't fix my problem so I'm posting my solution which was inspired by his suggestion.
I simply manually set a variable called title when creating the view.
def home():
return render_template(
'index.html',
title='Home',
)
Within the template I referenced the variable to decide which menu item to highlight.
{% if title == "Home" %}
<li class="active">
{% else %}
<li>
{% endif %}
It turned out in my situation I didn't need to access the current url.
Related
I have a function to check if the user is a staff:
class VerificateUser():
def allowed_user(request):
if request.user.is_staff:
return True
In my template, I’m going to show this section only for staff users, but this is not working.
{% url if not allowed_user %}
<h1>
Welcome to the show
</h1>
If do something like it, works:
```html
{% if not request.user.is_staff %}
<h1>
Welcome to the show
</h1>
But I need to use a view function to clean my code because I’m probably going to add more conditionals.
Since you are using a class-based view then I would suggest updating the context dictionary which you could use in the html template to determine whether the user is allowed or not. For example, Within the views.py.
class VerificateUser():
# Update the context
def get_context_data(self, **kwargs):
context = super().get_context_data(**kwargs)
# Call the allowed_user() and return whatever value, passing that value it to the context variable
context['allowed_user'] = self.allowed_user()
return context
# Checking if the user is allowed here
def allowed_user():
if self.request.user.is_staff:
return True
return False
Now within the html file, you can reference that allowed_user from the context variable.
{% if allowed_user %}
<h1>Hi, you are allowed to see here...</h1>
{% endif %}
That should do the trick.
You can do this sort of thing in many ways, but simply you can do by the following way-
{% if request.user.is_authenticated %}
<p>Welcome,{{request.user.first_name}} </p>
{% endif %}
request object is by default available in all of your Django templates context.
I need to apply a if statement in template file based on response code with respect to a url. As I'm not aware of any direct method, I'm planning to make a custom template tag like:
from django import template
register = template.Library()
#register.filter(name='code')
def code(url):
a = http.response(url)
return a
I will then call this within the template as:
{% if model.fileurl|code==200 %}
<div>..............</div>
Also, is there any way to do it directly within the template as I'm trying to fill in the shoes of our django developer.
You can make a template tag that returns the status code:
# app/templatetags/code.py
from django import template
from requests import get as reqget
register = template.Library()
#register.filter(name='code')
def code(url):
reqget(url).status_code
The app/templatetags/ directory need to contain an __init__.py file as well (if this is not the case yet, you can add an empty one).
and then in the template:
{% load code %}
{% if model.fileurl|code == 200 %}
<div>..............</div>
{% endif %}
EDITED. My original question wasn't clear enough. I want to allow a user to pass values into a TextField in wtforms, then the value they entered appears after they add it. This would allow the user to pass multiple values before then hitting a final "Sumbit" button on all the values that were originally entered.
I found this question for doing something with the entered text, which is what I tried.
My Python code:
from flask import Flask, request, render_template, redirect
from wtforms import TextField, Form, SubmitField
def redirect_url(default='index'):
return request.args.get('next') or \
request.referrer or \
url_for(default)
class RegionForm(Form):
field = TextField('Region')
Submit = SubmitField('AddRegion')
fieldList = []
def main():
app = Flask(__name__)
#app.route('/region/', methods=['GET'])
def region():
form = RegionForm(request.form)
return render_template("region.html",
form=form)
#app.route('/add/', methods=['POST'])
def add():
request.form['fieldList'].append(request.form['field'])
return redirect(redirect_url())
app.run(debug=True)
My html code:
<form action="/add/" method="Post">
{% for field in form %}
<tr>
<th>{{ field.label }}</th>
<td>{{ field }}</td>
</tr>
{% endfor %}
</form>
{% for item in form.fieldList %}
{{ item }}
{% endfor %}
But after I enter the text and click the "AddRegion" button, I get the following error: The browser (or proxy) sent a request that this server could not understand. However, if I comment out the line request.form['fieldList'].append(request.form['field']), then the redirect happens, but the text hasn't been added to the hidden list on the form. How do I both add the text to the list and redirect back to the original page, so the user can add more text? It seems like there must be an error with this line only, because the rest works fine.
How can I allow a user to dynamically add text to a field, then have that field display in the browser?
Then once the complete region fields have been added, I want to be able to retrieve that list to process in a separate function later.
Part One
So after looking at your code, I think I have found your problem. Flask is very particular about its app routes.
The app route that you have in your flask is:
#app.route('/add', methods=['POST'])
However, your current action on your form which is:
<form action="/add/" method="Post">
In flask /add and /add/ are actually two different web-routes. Therefore, the #app.route is not being triggered. If you change your code to:
`<form action="/add" method="post">`
You should be good to go from here.
Part Two
I think I may have an additional issue. So within your HTML right now, you actually close your </form> tag before looping through your items in the fieldList.
</form>
{% for item in form.fieldList %}
{{ item }}
{% endfor %}
Try:
{% for item in form.fieldList %}
{{ item }}
{% endfor %}
</form>
What I believe to be happening is that your form inputs are not actually being placed inside of the form so when you try to access them you are getting a KeyError.
I second what Cody Myers said. However there's a simple way to guarantee correct routes even if you later change them: in your form use action="{{ url_for('region') }}" and Flask will automatically substitute the correct route for the given function name.
I have the following code in my template:
{% for req in user.requests_made_set.all %}
{% if not req.is_published %}
{{ req }}
{% endif %}
{% empty %}
No requests
{% endfor %}
If there are some requests but none has the is_published = True then how could I output a message (like "No requests") ?? I'd only like to use Django templates and not do it in my view!
Thanks
Even if this might be possible to achieve in the template, I (and probably many other people) would advise against it. To achieve this, you basically need to find out whether there are any objects in the database matching some criteria. That is certainly not something that belongs into a template.
Templates are intended to be used to define how stuff is displayed. The task you're solving is determining what stuff to display. This definitely belongs in a view and not a template.
If you want to avoid placing it in a view just because you want the information to appear on each page, regardless of the view, consider using a context processor which would add the required information to your template context automatically, or writing a template tag that would solve this for you.
I have a form that redirects to the same page after a user enters information (so that they can continue entering information). If the form submission is successful, I'm returning
HttpResponseRedirect(request.path)
which works fine. However, I'd also like to display some messages to the user in this case (e.g., "Your data has been saved" at the top of the screen). If I weren't redirecting, I'd just return these messages in the context dictionary. With the redirect, however, I can't do this.
So how can I pass template context information when using HttpResponseRedirect?
What I'm trying to do seems like it would be incredibly common, so please excuse me if I'm missing something obvious.
For the sake of completion and future reference, you can now use the messages framework. After you install it:
views.py
from django.contrib import messages
def view(request):
# your code
messages.success(request, "Your data has been saved!")
HttpResponseRedirect(request.path)
template.html
{% if messages %}
<ul class="messages">
{% for message in messages %}
<li{% if message.tags %} class="{{ message.tags }}"{% endif %}>{{ message }}</li>
{% endfor %}
</ul>
{% endif %}
if you are using auth and have a logged in user you could:
http://docs.djangoproject.com/en/dev/topics/auth/#django.contrib.auth.models.User.message_set.create
GET params are also hackable. The querystring, as mentioned in other answers, could be used.
I think the most preferred way would be to use the sessions framework. That way you can load up whatever you want in the context and get
{{ request.session.foo }}
foo could be the message or you could do:
{% ifequal request.session.foo 1 %} Nice work! {% else %} Almost! {% endifequal %}
and other fun stuff.
http://docs.djangoproject.com/en/dev/topics/http/sessions/#using-sessions-in-views
You can't. HttpResponseRedirect sends a client-side redirect (HTTP status code 302) to the browser, and then the browser re-requests another page.
You can set a URL query string on the redirect, though that will be visible to the user and anyone intercepting HTTP requests (i.e. proxies), and is therefore not suitable for sensitive information.
The best way would probably be to use a coded querystring on the redirect URL... its an old school approach.
You could do something like
/page/?m=1, /page/?m=2, etc
You would then parse that variable with request.GET in the view code and show the appropriate message.
From your views.py you hast have to put a key/value-pair into the session and then read it from the HTML template.
For example:
views.py
# your code here
request.session['vote'] = 1
return HttpResponseRedirect(request.path)
your_template.html
{% ifequal request.session.vote 1 %}
<!-- Your action here -->
{% endifequal %}
The only way I know of to pass any data with a redirect is to add GET parameters to the URL you're passing in. To avoid XSS hacks you'd want to pass a specific constant like:
[current path you're passing in]?message=saved
And then process the message=saved parameter in the handler for the path you passed in.
A somewhat more complicated way would be not passing the data in the redirect, and instead using something like http://code.google.com/p/django-notify/ to store session-based data that is displayed to the user following the redirect.
You add ?saved=1 to the query string and check for it with something like:
saved = request.GET.get('saved', False)