How can I show a please wait loading message from a django view?
I have a Django view that takes significant time to perform calculations on a large dataset.
While the process loads, I would like to present the user with a feedback message e.g.: spinning loading animated gif or similar.
After trying the two different approaches suggested by Brandon and Murat, Brandon's suggestion proved the most successful.
Create a wrapper template that includes the javascript from http://djangosnippets.org/snippets/679/. The javascript has been modified: (i) to work without a form (ii) to hide the progress bar / display results when a 'done' flag is returned (iii) with the JSON update url pointing to the view described below
Move the slow loading function to a thread. This thread will be passed a cache key and will be responsible for updating the cache with progress status and then its results. The thread renders the original template as a string and saves it to the cache.
Create a view based on upload_progress from http://djangosnippets.org/snippets/678/ modified to (i) instead render the original wrapper template if progress_id='' (ii) generate the cache_key, check if a cache already exists and if not start a new thread (iii) monitor the progress of the thread and when done, pass the results to the wrapper template
The wrapper template displays the results via document.getElementById('main').innerHTML=data.result
(* looking at whether step 4 might be better implemented via a redirect as the rendered template contains javascript that is not currently run by document.getElementById('main').innerHTML=data.result)
Another thing you could do is add a javascript function that displays a loading image before it actually calls the Django View.
function showLoaderOnClick(url) {
showLoader();
window.location=url;
}
function showLoader(){
$('body').append('<div style="" id="loadingDiv"><div class="loader">Loading...</div></div>');
}
And then in your template you can do:
This will take some time...
Here's a quick default loadingDiv : https://stackoverflow.com/a/41730965/13476073
Note that this requires jQuery.
a more straightforward approach is to generate a wait page with your gif etc. and then use the javascript
window.location.href = 'insert results view here';
to switch to the results view which starts your lengthy calculation. The page wont change until the calculation is finished. When it finishes, then the results page will be rendered.
Here's an oldie, but might get you going in the right direction: http://djangosnippets.org/snippets/679/
A workaround that I chose was to use beforunload and unload events to show the loading image. This can be used with or without window.load. In my case, it's the view that is taking a great amount of time and not the page loading, hence I am not using window.load (because it's already a lot of time by the time window.load comes into picture, and at that point of time, I do not need the loading icon to be shown anymore).
The downside is that there is a false message that goes out to the user that the page is loading even when when the request has not even reached the server or it's taking much time. Also, it doesn't work for requests coming from outside my website. But I'm living with this for now.
Update: Sorry for not adding code snippet earlier, thanks #blockhead. The following is a quick and dirty mix of normal JS and JQuery that I have in the master template.
Update 2: I later moved to making my view(s) lightweight which send the crucial part of the page quickly, and then using ajax to get the remaining content while showing the loading icon. It needed quite some work, but the end result is worth it.
window.onload=function(){
$("#load-icon").hide(); // I needed the loading icon to hide once the page loads
}
var onBeforeUnLoadEvent = false;
window.onunload = window.onbeforeunload= function(){
if(!onBeforeUnLoadEvent){ // for avoiding dual calls in browsers that support both events
onBeforeUnLoadEvent = true;
$("#load-icon").show();
setTimeout(function(){
$("#load-icon").hide();},5000); // hiding the loading icon in any case after
// 5 seconds (remove if you do not want it)
}
};
P.S. I cannot comment yet hence posted this as an answer.
Iterating HttpResponse
https://stackoverflow.com/a/1371061/198062
Edit:
I found an example to sending big files with django: http://djangosnippets.org/snippets/365/ Then I look at FileWrapper class(django.core.servers.basehttp):
class FileWrapper(object):
"""Wrapper to convert file-like objects to iterables"""
def __init__(self, filelike, blksize=8192):
self.filelike = filelike
self.blksize = blksize
if hasattr(filelike,'close'):
self.close = filelike.close
def __getitem__(self,key):
data = self.filelike.read(self.blksize)
if data:
return data
raise IndexError
def __iter__(self):
return self
def next(self):
data = self.filelike.read(self.blksize)
if data:
return data
raise StopIteration
I think we can make a iterable class like this
class FlushContent(object):
def __init__(self):
# some initialization code
def __getitem__(self,key):
# send a part of html
def __iter__(self):
return self
def next(self):
# do some work
# return some html code
if finished:
raise StopIteration
then in views.py
def long_work(request):
flushcontent = FlushContent()
return HttpResponse(flushcontent)
Edit:
Example code, still not working:
class FlushContent(object):
def __init__(self):
self.stop_index=2
self.index=0
def __getitem__(self,key):
pass
def __iter__(self):
return self
def next(self):
if self.index==0:
html="loading"
elif self.index==1:
import time
time.sleep(5)
html="finished loading"
self.index+=1
if self.index>self.stop_index:
raise StopIteration
return html
Here is another explanation on how to get a loading message for long loading Django views
Views that do a lot of processing (e.g. complex queries with many objects, accessing 3rd party APIs) can take quite some time before the page is loaded and shown to the user in the browser. What happens is that all that processing is done on the server and Django is not able to serve the page before it is completed.
The only way to show a show a loading message (e.g. a spinner gif) during the processing is to break up the current view into two views:
First view renders the page with no processing and with the loading message
The page includes a AJAX call to the 2nd view that does the actual processing. The result of the processing is displayed on the page once its done with AJAX / JavaScript
Related
I want an effect to be applied when a user is entering my website. So therefore I want to check for when a user is coming from outside my website so the effect isnt getting applied when the user is surfing through different urls inside the website, but only when the user is coming from outside my website
You can't really check for where a user has come from specifically. You can check if the user has just arrived on your site by setting a session variable when they load one of your pages. You can check for it before you set it, and if they don't have it, then they have just arrived and you can apply your effect. There's some good examples of how sessions work here: https://developer.mozilla.org/en-US/docs/Learn/Server-side/Django/Sessions
There's a couple of ways to handle this. If you are using function based views, you can just create a separate util function and include it at the top of every page, eg,
utils.py
def first_visit(request):
"""returns the answer to the question 'first visit for session?'
make sure SESSION_EXPIRE_AT_BROWSER_CLOSE set to False in settings for persistance"""
if request.session['first_visit']:
#this is not the first session because the session variable is used.
return False
else:
#This is the first visit
...#do something
#set the session variable so you only do the above once
request.session[first_visit'] = True
return True
views.py
from utils.py import first_visit
def show_page(request):
first_visit = first_visit(request)
This approach gives you some control. For example, you may not want to run it on pages that require login, because you will already have run it on the login page.
Otherwise, the best approach depends on what will happen on the first visit. If you want just to update a template (eg, perhaps to show a message or run a script on th epage) you can use a context processor which gives you extra context for your templates. If you want to interrupt the request, perhaps to redirect it to a separate page, you can create a simple piece of middleware.
docs for middleware
docs for context processors
You may also be able to handle this entirely by javascript. This uses localStorage to store whether or not this is the user's first visit to the site and displays the loading area for 5 seconds if there is nothing in localStorage. You can include this in your base template so it runs on every page.
function showMain() {
document.getElementByID("loading").style.display = "none";
document.getElementByID("main").style.display = "block";
}
const secondVisit = localStorage.getItem("secondVisit");
if (!secondVisit) {
//show loading screen
document.getElementByID("loading").style.display = "block";
document.getElementByID("main").style.display = "none";
setTimeout(5000, showMain)
localStorage.setItem("secondVisit", "true" );
} else {
showMain()
}
I need to scrap the data of each item from a website using Scrapy(http://example.com/itemview). I have a list of itemID and I need to pass it in a form in example.com.
There is no url change for each item. So for each request in my spider the url will always be the same. But the content will be different.
I don't wan't a for loop for handling each request. So i followed the below mentioned steps.
started spider with the above url
added item_scraped and spider_closed signals
passed through several functions
passed the scraped data to pipeline
trigerred the item_scraped signal
After this it automatically calls the spider_closed signal. But I want the above steps to be continued till the total itemID are finished.
class ExampleSpider(scrapy.Spider):
name = "example"
allowed_domains = ["example.com"]
itemIDs = [11111,22222,33333]
current_item_num = 0
def __init__(self, itemids=None, *args, **kwargs):
super(ExampleSpider, self).__init__(*args, **kwargs)
dispatcher.connect(self.item_scraped, signals.item_scraped)
dispatcher.connect(self.spider_closed, signals.spider_closed)
def spider_closed(self, spider):
self.driver.quit()
def start_requests(self):
request = self.make_requests_from_url('http://example.com/itemview')
yield request
def parse(self,response):
self.driver = webdriver.PhantomJS()
self.driver.get(response.url)
first_data = self.driver.find_element_by_xpath('//div[#id="itemview"]').text.strip()
yield Request(response.url,meta={'first_data':first_data},callback=self.processDetails,dont_filter=True)
def processDetails(self,response):
itemID = self.itemIDs[self.current_item_num]
..form submission with the current itemID goes here...
...the content of the page is updated with the given itemID...
yield Request(response.url,meta={'first_data':response.meta['first_data']},callback=self.processData,dont_filter=True)
def processData(self,response):
...some more scraping goes here...
item = ExamplecrawlerItem()
item['first_data'] = response.meta['first_data']
yield item
def item_scraped(self,item,response,spider):
self.current_item_num += 1
#i need to call the processDetails function here for the next itemID
#and the process needs to contine till the itemID finishes
self.parse(response)
My piepline:
class ExampleDBPipeline(object):
def process_item(self, item, spider):
MYCOLLECTION.insert(dict(item))
return
I wish I had an elegant solution to this. But instead it's a hackish way of calling the underlying classes.
self.crawler.engine.slot.scheduler.enqueue_request(scrapy.Request(url,self.yourCallBack))
However, you can yield a request after you yield the item and have it callback to self.processDetails. Simply add this to your processData function:
yield item
self.counter += 1
yield scrapy.Request(response.url,callback=self.processDetails,dont_filter=True, meta = {"your":"Dictionary"}
Also, PhantomJS can be nice and make your life easy, but it is slower than regular connections. If possible, find the request for json data or whatever makes the page unparseable without JS. To do so, open up chrome, right click, click inspect, go to the network tab, then enter the ID into the form, then look at the XHR or JS tabs for a JSON that has the data or next url you want. Most of the time, there will be some url made by adding the ID, if you can find it, you can just concatenate your urls and call that directly without having the cost of JS rendering. Sometimes it is randomized, or not there, but I've had fair success with it. You can then also use that to yield many requests at the same time without having to worry about phantomJS trying to do two things at once or having to initialize many instances of it. You could use tabs, but that is a pain.
Also, I would use a Queue of your IDs to ensure thread safety. Otherwise, you could have processDetails called twice on the same ID, though in the logic of your program everything seems to go linearly, which means you aren't using the concurrency capabilities of Scrapy and your program will go more slowly. To use Queue add:
import Queue
#go inside class definition and add
itemIDQueue = Queue.Queue()
#within __init__ add
[self.itemIDQueue.put(ID) for ID in self.itemID]
#within processDetails replace itemID = self.itemIDs[self.current_item_num] with
itemID = self.itemIDQueue.get()
And then there is no need to increment the counter and your program is thread safe.
I'm very new to programming and especially to Django but can't work out how to use any previous answers to my advantage....
Apologies if my question is too vague but essentially, I have two different apps, let's call them app A and app B, with data on two different databases but apps contain information on the same individual item.
I want to edit this information on my 'edit details' page while keeping the apps as separate as possible (well AppB can know about functions in AppA but not vice-versa)...I guess what I really want is a signal which works like so:
A 'submit' view within AppA which is called when I submit changes to the data (using text boxes). The data for AppA is then saved..
And a signal then sent to AppB which ideally would update its data, before the HttpResponseRedirect is performed.
Unfortunately I can't really get this to work. My problem is that if I put 'request' into the arguments for save_details, I get errors like "save_details() takes exactly 3 arguments (2 given)"....does anyone know a clever way of getting something like this to work?
My submit function in AppA looks something like this...
def submit(self, request, id):
signal_received.send(sender=self, id=id)
q = get_object_or_404(AppA, pk=id)
q.blah = request.POST.get('wibble from the form')
...
return Http.....
in my AppB signals.py file, I have put.
signal_received = django.dispatch.Signal(providing_args=['id'])
def save_details(sender, uid, **kwargs):
p = AppB.objects.get(id=id)
p.wobble = request.POST.get('wobble from the form')
...
signal.received.connect(save_details)
Obviously the above doesn't mention request in its arguments which seems to be necessary but if I add that, I get problems with the number of arguments.
(I have imported all the right things at the top of each file I think...hence me leaving that off.)
Any point about the above would be appreciated....e.g. does "request" need to be the first argument? It didn't seem to like me using "self" before but I have tried to copy as much as possible the example at the bottom of the documentation (https://docs.djangoproject.com/en/dev/topics/signals/) but the extra functionality I need in the signal receiving function is flumoxing me.
Thanks in advance...
I have a following situation. I created a simple backend in Flask that handles file uploads. With files received, Flask does something (uploads them), and returns the data to the caller. There are two scenarios with the app, to upload one image and multiple images. When uploading one image, I can simply get the response and voila, I'm all set.
However, I am stuck on handling multiple file uploads. I can use the same handler for the actual file upload, but the issue is that all of those files need to be stored into a list or something, then processed, and after doing that, a single link (album) containing all those images, needs to be delivered.
Here is my upload handling code:
#app.route('/uploadv3', methods=['POST'])
def upload():
if request.method == 'POST':
data_file = request.files["file"]
file_name = data_file.filename
path_to_save_to = os.path.join(app.config['UPLOAD_FOLDER'], file_name)
data_file.save(path_to_save_to)
file_url = upload_image_to_image_host(path_to_save_to)
return file_url
I was experimenting with session in flask, but I dont know can I create a list of items under one key, like session['links'], and then get all those, and clear it after doing the work. Or is there some other simpler solution?
I assume that I could probably do this via key for each image, like session["link1"], and so on, but that would impose a limit on the images (depending on how much of those I create), would make the code very ugly, make the iteration over each in order to generate a list that is passed to an album building method problematic, and session clearing would be tedious.
Some code that I wrote for getting the actual link at the end and clearing the session follows (this assume that session['link'] has a list of urls, which I can't really achieve with my knowledge of session management in Flask:
def create_album(images):
session.pop('link', None)
new_album = im.create_album(images)
return new_album.link
#app.route('/get_album_link')
def get_album_link():
return create_album(session['link'])
Thanks in advance for your time!
You can assign anything to a session including individual value or list/dictionary etc. If you know the links, you can store them in the session as follows:
session['links'] = ['link1','link2'...and so on]
This way, you have a list of all the links. You can now access a link by:
if 'links' in session:
for link in session['links']:
print link
Once you are done with them, you can clear the session as:
if 'links' in session:
del session['links']
To clarify what I have done to make this work. At the end, it appeared that the uploading images and adding them to the album anonymously had to be done "reversely", so not adding images to an album object, but uploading an image object to an album id.
I made a method that gets the album link and puts it in the session:
#app.route('/get_album_link')
def get_album_link():
im = pyimgur.Imgur(CLIENT_ID)
new_album = im.create_album()
session.clear()
session['album'] = new_album.deletehash
session['album_link'] = new_album.link
return new_album.link
Later on, when handling uploads, I just add the image to the album and voila, all set :)
uploaded_image = im.upload_image(path_of_saved_image, album=session['album'])
file_url = uploaded_image.link
return file_url
One caveat is that the image should be added to the "deleteahash" value passed as the album value, not the album ID (which is covered by the imgur api documentation).
I'm trying to call a view directly from another (if this is at all possible). I have a view:
def product_add(request, order_id=None):
# Works. Handles a normal POST check and form submission and redirects
# to another page if the form is properly validated.
Then I have a 2nd view, that queries the DB for the product data and should call the first one.
def product_copy_from_history(request, order_id=None, product_id=None):
product = Product.objects.get(owner=request.user, pk=product_id)
# I need to somehow setup a form with the product data so that the first
# view thinks it gets a post request.
2nd_response = product_add(request, order_id)
return 2nd_response
Since the second one needs to add the product as the first view does it I was wondering if I could just call the first view from the second one.
What I'm aiming for is just passing through the request object to the second view and return the obtained response object in turn back to the client.
Any help greatly appreciated, critism as well if this is a bad way to do it. But then some pointers .. to avoid DRY-ing.
Thanx!
Gerard.
My god, what was I thinking. This would be the cleanest solution ofcourse:
def product_add_from_history(request, order_id=None, product_id=None):
""" Add existing product to current order
"""
order = get_object_or_404(Order, pk=order_id, owner=request.user)
product = Product.objects.get(owner=request.user, pk=product_id)
newproduct = Product(
owner=request.user,
order = order,
name = product.name,
amount = product.amount,
unit_price = product.unit_price,
)
newproduct.save()
return HttpResponseRedirect(reverse('order-detail', args=[order_id]) )
A view is a regular python method, you can of course call one from another giving you pass proper arguments and handle the result correctly (like 404...). Now if it is a good practice I don't know. I would myself to an utiliy method and call it from both views.
If you are fine with the overhead of calling your API through HTTP you can use urllib to post a request to your product_add request handler.
As far as I know this could add some troubles if you develop with the dev server that comes with django, as it only handles one request at a time and will block indefinitely (see trac, google groups).