I have a function calculate_full_eva_web(input:dict) it receives input dictionary several function applied on this input to create calculations dict, after calculations i want to send this data to html dashboard and after send data to html file i can play there with jinja stuff. i am unable to do so, i tried several ways but flask throws error. and also i don't know much about ajax ,may be ajax will do my work, let me know. that is why i am tagging ajax people on this post. Traceback is also attached..Thank you
In simple words, i want to send data to html in flask ! Please check my code. Let me know if i am doing anything wrong.
imports ...
from other file import other_functions
from other file import other_functions_2
from other file import other_functions_3
app = Flask(__name__, template_folder='templates/')
#app.route("/dashboard")
def calculate_full_eva_web(input:dict):
calculate_gap = other_functions(input)
calculate_matrix = other_functions_2(input)
average = other_functions_3(input)
data = dict{'calculate_gap':calculate_gap, 'calculate_matrix':calculate_matrix,'average':average}
return render_template('pages/dashboard.html', data = data)
if __name__ == "__main__":
app.run(debug=True)
The route receive a dict as input so you must change #app.route("/dashboard") to #app.route("/dashboard/<input>") and pass input to the route in the link of the route.
For example, I have a route as below.
#app.route('/user/<name>')
def user(name):
return render_template('home.html', name=name)
To pass name to the route, I access the link http://localhost:5000/user/myname.
Related
Question
I have created a dynamic route as /update/<randomString> in my Flask app.py file, where randomString is a randomly generated string with the length of 50. However if I search for /update/1 I am able to view the same dynamic route /update/<randomString> without any error! Can anyone explain why is it so?
See what I've tried so far:
#app.route('/')
def index():
randomString = ''.join(secrets.choice(string.ascii_uppercase+string.digits+string.ascii_lowercase) for k in range (50))
session['randomString'] = str(randomString)
return render_template('index.html')
#app.route('/update/<randomString>')
def update(randomString):
if 'randomString' in session:
randomString = session['randomString']
return render_template('update.html')
else:
return 'error...'
Link of the dynamic page at update.html page, where random string is passed with the help of session(defined at index.html page).
Dynamic page
Edit: I am also able to view dynamic route when I click on the link defined above and my URL section shows that long randomString. Problem is: I can access the same route when I search for http://127.0.0.1:5000/update/1
Screenshot one
Screenshot two
While storing the random string, the key you use is randomString. So you are storing the random string in a dict like
session['randomString'] = '1234567890'
Then when you access the session in the /update route you are just checking if session has a key named randomString. You should also check if session['randomString'] == '1234567890' and render the page only if the random string in session is the same as you created in the / path. You can replace the if with
if 'randomString' in session and session['randomString'] == randomString :
I stuck on the problem of how to organize code / proper way to get xlsx file as output in flask app.
I have a function.py file where for now the xlsx file generates.
The sense is that flask app gets some settings in json format, these settings are processed by function that must return xlsx(?) to app.
The function do some calculations depending on the settings.
The file has the next structure:
def function (settings):
settings=settings
df = pd.read_csv(settings['df'])
from pandas import ExcelWriter
writer = ExcelWriter('file.xlsx')
if settings[somefeature1]==1:
f1=dosmth.to_excel(writer, "feature 1")
if settings[somefeature2]==1:
f2=dosmth.to_excel(writer, "feature 2")
...
writer.save()
But if the file is already generated in function, what should I pass to flask? How the flask app function must look like then (especially in case if I want to return xlsx as json)?
#app.route('/Function', methods = ['POST'])
def Function():
settings = request.get_json(force = True)
return(function(settings)) #???
You should never forget that flask is a framework for creating web applications, a web application is a piece of software that receives a web request and generates a web response.
To make it super simple: your flask function should return something that a common web browser will be able to handle.
In this case your response should return the file but also some metadata to tell to the potential receiver what is inside the response and how to handle it.
I think that something like this could work:
return send_file(filename, mimetype='application/vnd.ms-excel')
first off, a disclaimer: I'm not well versed in python or flask, so bear with me.
I'm trying to put together a minimal API using flask, i was planning to dynamically generate routes and their associated procs from the contents of a subdirectory.
The code looks something like this:
from flask import Flask
import os
app = Flask(__name__)
configs = os.getcwd() + "/configs"
for i in os.listdir(configs):
if i.endswith(".json"):
call = "/" + os.path.splitext(i)[0]
#app.route(call, methods=['POST'])
def call():
return jsonify({"status": call + "Success"}), 200
The plan being to iterate over a bunch of config files and use their naes to define the routes. Now, this works for a single config file, but wont work for multiple files as I end up trying to overwrite the function call that is used by each route.
I can factor out most of the code to a separate function as long as i can pass in the call name. However it seems that however i go about this i need to dynamically name the function generated and mapped to the route.
So, my question is: how can use the contents of a variable, such as 'call' to be the function name?
i.e. something like
call = "getinfo"
def call(): # Effectively being evaled as def getinfo():
Everything i've tried hasn't worked, and i'm not confident enough in my python syntax to know if it's because i'm just doing something silly.
Alternatively is there another way to do what i'm trying to achieve?
Thanks for all and any feedback!
Thanks for the help. I've moved to one route and one handler and building up the file list, and handling of the request paths, etc separately.
This is a sanitized version of the model i now have:
from flask import Flask
import os
calls = []
cfgs = {}
app = Flask(__name__)
configs = os.getcwd() + "/configs"
for i in os.listdir(configs):
if i.endswith(".json"):
cfgs[call] = os.path.splitext(i)[0]
calls.extend([call])
#app.route('/<call>', methods=['POST'])
def do(call):
if call not in calls:
abort(400, "invalid call")
# Do stuff
return jsonify({"status": call + "Success"}), 200
if __name__ == '__main__':
app.run(debug=True)
So, thanks to the above comments this is doing what i'm after. Still curious to know if there is any way to use variables in function names?
So, I'm trying to make a simple call using jQuery .getJSON to my local web server using python/django to serve up its requests. The address being used is:
http://localhost:8000/api/0.1/tonight-mobile.json?callback=jsonp1290277462296
I'm trying to write a simple web view that can access this url and return a JSON packet as the result (worried about actual element values/layout later).
Here's my simple attempt at just alerting/returning the data:
$.getJSON("http://localhost:8000/api/0.1/tonight-mobile.json&callback=?",
function(json){
alert(json);
<!--$.each(json.items, function(i,item){
});-->
});
I am able to access this URL directly, either at http://localhost:8000/api/0.1/tonight-mobile.json or http://localhost:8000/api/0.1/tonight-mobile.json&callback=jsonp1290277462296 and get back a valid JSON packet... So I'm assuming it's in my noob javascript:)
My views.py function that is generating this response looks as follows:
def tonight_mobile(request):
callback = request.GET.get('callback=?', '')
def with_rank(rank, place):
return (rank > 0)
place_data = dict(
Places = [make_mobile_place_dict(request, p) for p in Place.objects.all()]
)
xml_bytes = json.dumps(place_data)
xml_bytes = callback + '(' + xml_bytes + ');'
return HttpResponse(xml_bytes, mimetype="application/json")
With corresponding urls.py configuration:
(r'^tonight-mobile.json','iphone_api.views.tonight_mobile'),
I am still somewhat confused on how to use callbacks, so maybe that is where my issue lies. Note I am able to call directly a 'blah.json' file that is giving me a response, but not through a wired URL. Could someone assist me with some direction?
First, callback = request.GET.get('callback=?', '') won't get you the value of callback.
callback = request.GET.get( 'callback', None )
Works much better.
To debug this kind of thing. You might want to include print statements in your Django view function so you can see what's going on. For example: print repr(request.GET) is a helpful thing to put in a view function so that you can see the GET dictionary.
I am trying to have a custom URL which looks like this:
example.com/site/yahoo.com
which would hit this script like this=
example.com/details?domain=yahoo.com
can this be done using app.yaml?
the basic idea is to call "details" with the input "yahoo.com"
You can't really rewrite the URLs per se, but you can use regular expression groups to perform a similar kind of thing.
In your app.yaml file, try something like:
handlers:
- url: /site/(.+)
script: site.py
And in your site.py:
SiteHandler(webapp.RequestHandler):
def get(self, site):
# the site parameter will be what was passed in the URL!
pass
def main():
application = webapp.WSGIApplication([('/site/(.+)', SiteHandler)], debug=True)
util.run_wsgi_app(application)
What happens is, whatever you have after /site/ in the request URL will be passed to SiteHandler's get() method in the site parameter. From there you can do whatever it is you wanted to do at /details?domain=yahoo.com, or simply redirect to that URL.