I'm making a REST api that files can be uploaded based in MODEL-VIEW in flask-appbuilder like this.
But I don't know how to call REST API (POST /File).
I tried several different ways. but I couldn't.
Let me know the correct or the alternative ways.
[client code]
file = {'file':open('test.txt', 'rb'),'description':'test'}
requests.post(url, headers=headers, files=file)
==> Failed
model.py
class Files(Model):
__tablename__ = "project_files"
id = Column(Integer, primary_key=True)
file = Column(FileColumn, nullable=False)
description = Column(String(150))
def download(self):
return Markup(
'<a href="'
+ url_for("ProjectFilesModelView.download", filename=str(self.file))
+ '">Download</a>'
)
def file_name(self):
return get_file_original_name(str(self.file))
view.py
class FileApi(ModelRestApi):
resource_name = "File"
datamodel = SQLAInterface(Files)
allow_browser_login = True
appbuilder.add_api(FileApi)
FileColumn is only a string field that saves the file name in the database. The actual file is saved to config['UPLOAD_FOLDER'].
This is taken care of by flask_appbuilder.filemanager.FileManager.
Furthermore, ModelRestApi assumes that you are POSTing JSON data. In order to upload files, I followed Flask's documentation, which suggests to send a multipart/form-data request. Because of this, one needs to override ModelRestApi.post_headless().
This is my solution, where I also make sure that when a Files database row
is deleted, so is the relative file from the filesystem.
from flask_appbuilder.models.sqla.interface import SQLAInterface
from flask_appbuilder.api import ModelRestApi
from flask_appbuilder.const import API_RESULT_RES_KEY
from flask_appbuilder.filemanager import FileManager
from flask import current_app, request
from marshmallow import ValidationError
from sqlalchemy.exc import IntegrityError
from app.models import Files
class FileApi(ModelRestApi):
resource_name = "file"
datamodel = SQLAInterface(Files)
def post_headless(self):
if not request.form or not request.files:
msg = "No data"
current_app.logger.error(msg)
return self.response_400(message=msg)
file_obj = request.files.getlist('file')
if len(file_obj) != 1:
msg = ("More than one file provided.\n"
"Please upload exactly one file at a time")
current_app.logger.error(msg)
return self.response_422(message=msg)
else:
file_obj = file_obj[0]
fm = FileManager()
uuid_filename = fm.generate_name(file_obj.filename, file_obj)
form = request.form.to_dict(flat=True)
# Add the unique filename provided by FileManager, which will
# be saved to the database. The original filename can be
# retrieved using
# flask_appbuilder.filemanager.get_file_original_name()
form['file'] = uuid_filename
try:
item = self.add_model_schema.load(
form,
session=self.datamodel.session)
except ValidationError as err:
current_app.logger.error(err)
return self.response_422(message=err.messages)
# Save file to filesystem
fm.save_file(file_obj, item.file)
try:
self.datamodel.add(item, raise_exception=True)
return self.response(
201,
**{API_RESULT_RES_KEY: self.add_model_schema.dump(
item, many=False),
"id": self.datamodel.get_pk_value(item),
},
)
except IntegrityError as e:
# Delete file from filesystem if the db record cannot be
# created
fm.delete_file(item.file)
current_app.logger.error(e)
return self.response_422(message=str(e.orig))
def pre_delete(self, item):
"""
Delete file from filesystem before removing the record from the
database
"""
fm = FileManager()
current_app.logger.info(f"Deleting {item.file} from filesystem")
fm.delete_file(item.file)
You can use this.
from app.models import Project, ProjectFiles
class DataFilesModelView(ModelView):
datamodel = SQLAInterface(ProjectFiles)
label_columns = {"file_name": "File Name", "download": "Download"}
add_columns = ["file", "description", "project"]
edit_columns = ["file", "description", "project"]
list_columns = ["file_name", "download"]
show_columns = ["file_name", "download"]
Last add the view to the menu.
appbuilder.add_view(DataFilesModelView,"File View")
am new in python and i made a program to add data from a form to database am using wamp server here is my app.py code
import web
import MySQLdb
urls = (
'/', 'Index'
)
app = web.application(urls, globals())
render = web.template.render('templates/', base = "layout")
class Index(object):
def GET(self):
return render.hello_form()
def POST(self): #Open database connection
try:
db = MySQLdb.connect("localhost", "root", "", "testdb")
# prepare a cursor object using cursor() method
cursor = db.cursor()
# Prepare SQL query to INSERT a record into the database.
sql = "INSERT INTO details(name, address,) VALUES ("[self.name.text, self.address.text])""
try:
# Execute the SQL command
cursor.execute(sql)
# Commit your changes in the database
db.commit()
except:
# Rollback in case there is any error
db.rollback()
# disconnect from server
db.close()
finally:
form = web.input(name = "Nobody", greet = "Hello")
greeting = "%s, %s" % (form.name, form.address)
return render.index(greeting = greeting)
if __name__ == '__main__':
main()
and my error is Expected an indended block at line 12(return.render.hello_form)
can any one help pls..?
Scope is determined in python by indentation. As a result, you need to indent the try statement as it is inside the function POST. Also, the formatting on finally (and db.close()) looks wrong. Check out the this SO question, this guide and the Pep 8 guide on best practice code formatting. Also, be careful not to mix spaces and tabs as this can cause problems.
The following code is written using selenium python web driver which is run in saucelabs.I am providing the browser name,version and platform in a list,how do i do the same by providing the browser details through command line arguments? I am using py.test to execute the test cases.
import os
import sys
import httplib
import base64
import json
import new
import unittest
import sauceclient
from selenium import webdriver
from sauceclient import SauceClient
# it's best to remove the hardcoded defaults and always get these values
# from environment variables
USERNAME = os.environ.get('SAUCE_USERNAME', "ranjanprabhub")
ACCESS_KEY = os.environ.get('SAUCE_ACCESS_KEY', "ecec4dd0-d8da-49b9-b719-17e2c43d0165")
sauce = SauceClient(USERNAME, ACCESS_KEY)
browsers = [{"platform": "Mac OS X 10.9",
"browserName": "chrome",
"version": ""},
]
def on_platforms(platforms):
def decorator(base_class):
module = sys.modules[base_class.__module__].__dict__
for i, platform in enumerate(platforms):
d = dict(base_class.__dict__)
d['desired_capabilities'] = platform
name = "%s_%s" % (base_class.__name__, i + 1)
module[name] = new.classobj(name, (base_class,), d)
return decorator
#on_platforms(browsers)
class SauceSampleTest(unittest.TestCase):
def setUp(self):
self.desired_capabilities['name'] = self.id()
sauce_url = "http://%s:%s#ondemand.saucelabs.com:80/wd/hub"
self.driver = webdriver.Remote(
desired_capabilities=self.desired_capabilities,
command_executor=sauce_url % (USERNAME, ACCESS_KEY)
)
self.driver.implicitly_wait(30)
def test_sauce(self):
self.driver.get('http://saucelabs.com/test/guinea-pig')
assert "I am a page title - Sauce Labs" in self.driver.title
comments = self.driver.find_element_by_id('comments')
comments.send_keys('Hello! I am some example comments.'
' I should be in the page after submitting the form')
self.driver.find_element_by_id('submit').click()
commented = self.driver.find_element_by_id('your_comments')
assert ('Your comments: Hello! I am some example comments.'
' I should be in the page after submitting the form'
in commented.text)
body = self.driver.find_element_by_xpath('//body')
assert 'I am some other page content' not in body.text
self.driver.find_elements_by_link_text('i am a link')[0].click()
body = self.driver.find_element_by_xpath('//body')
assert 'I am some other page content' in body.text
def tearDown(self):
print("Link to your job: https://saucelabs.com/jobs/%s" % self.driver.session_id)
try:
if sys.exc_info() == (None, None, None):
sauce.jobs.update_job(self.driver.session_id, passed=True)
else:
sauce.jobs.update_job(self.driver.session_id, passed=False)
finally:
self.driver.quit()
So this is a bit complicated because you can pass an array of browsers into the #on_platforms decorator. My solution will only work for a single browser, as it looks like that's what you're doing right now.
For the current, single browser, situation -- you're looking for argparse. Here's my suggested fix:
import argparse
def setup_parser():
parser = argparse.ArgumentParser(description='Automation Testing!')
parser.add_argument('-p', '--platform', help='Platform for desired_caps', default='Mac OS X 10.9')
parser.add_argument('-b', '--browser-name', help='Browser Name for desired_caps', default='chrome')
parser.add_argument('-v', '--version', default='')
args = vars(parser.parse_args())
return args
desired_caps = setup_parser()
browsers = [desired_caps]
print browsers
But if you're looking to test multiple browsers (which I suggest you do!), you should not try and use command line arguments for the desired_caps of each individual browser. You should instead load a json config file for the browsers and the desired_caps for each one that you want Sauce to run.
Maybe have a different config file for each set of browsers, and then use command line arguments to pass in the config files you want to load.
pyGTrends does not seem to work. Giving errors in Python.
pyGoogleTrendsCsvDownloader seems to work, logs in, but after getting 1-3 requests (per day!) complains about exhausted quota, even though manual download with the same login/IP works flawlessly.
Bottom line: neither work. Searching through stackoverflow: many questions from people trying to pull csv's from Google, but no workable solution I could find...
Thank you in advance: whoever will be able to help. How should the code be changed? Do you know of another solution that works?
Here's the code of pyGoogleTrendsCsvDownloader.py
import httplib
import urllib
import urllib2
import re
import csv
import lxml.etree as etree
import lxml.html as html
import traceback
import gzip
import random
import time
import sys
from cookielib import Cookie, CookieJar
from StringIO import StringIO
class pyGoogleTrendsCsvDownloader(object):
'''
Google Trends Downloader
Recommended usage:
from pyGoogleTrendsCsvDownloader import pyGoogleTrendsCsvDownloader
r = pyGoogleTrendsCsvDownloader(username, password)
r.get_csv(cat='0-958', geo='US-ME-500')
'''
def __init__(self, username, password):
'''
Provide login and password to be used to connect to Google Trends
All immutable system variables are also defined here
'''
# The amount of time (in secs) that the script should wait before making a request.
# This can be used to throttle the downloading speed to avoid hitting servers too hard.
# It is further randomized.
self.download_delay = 0.25
self.service = "trendspro"
self.url_service = "http://www.google.com/trends/"
self.url_download = self.url_service + "trendsReport?"
self.login_params = {}
# These headers are necessary, otherwise Google will flag the request at your account level
self.headers = [('User-Agent', 'Mozilla/5.0 (Windows NT 6.1; WOW64; rv:12.0) Gecko/20100101 Firefox/12.0'),
("Accept", "text/html,application/xhtml+xml,application/xml;q=0.9,*/*;q=0.8"),
("Accept-Language", "en-gb,en;q=0.5"),
("Accept-Encoding", "gzip, deflate"),
("Connection", "keep-alive")]
self.url_login = 'https://accounts.google.com/ServiceLogin?service='+self.service+'&passive=1209600&continue='+self.url_service+'&followup='+self.url_service
self.url_authenticate = 'https://accounts.google.com/accounts/ServiceLoginAuth'
self.header_dictionary = {}
self._authenticate(username, password)
def _authenticate(self, username, password):
'''
Authenticate to Google:
1 - make a GET request to the Login webpage so we can get the login form
2 - make a POST request with email, password and login form input values
'''
# Make sure we get CSV results in English
ck = Cookie(version=0, name='I4SUserLocale', value='en_US', port=None, port_specified=False, domain='www.google.com', domain_specified=False,domain_initial_dot=False, path='/trends', path_specified=True, secure=False, expires=None, discard=False, comment=None, comment_url=None, rest=None)
self.cj = CookieJar()
self.cj.set_cookie(ck)
self.opener = urllib2.build_opener(urllib2.HTTPCookieProcessor(self.cj))
self.opener.addheaders = self.headers
# Get all of the login form input values
find_inputs = etree.XPath("//form[#id='gaia_loginform']//input")
try:
#
resp = self.opener.open(self.url_login)
if resp.info().get('Content-Encoding') == 'gzip':
buf = StringIO( resp.read())
f = gzip.GzipFile(fileobj=buf)
data = f.read()
else:
data = resp.read()
xmlTree = etree.fromstring(data, parser=html.HTMLParser(recover=True, remove_comments=True))
for input in find_inputs(xmlTree):
name = input.get('name')
if name:
name = name.encode('utf8')
value = input.get('value', '').encode('utf8')
self.login_params[name] = value
except:
print("Exception while parsing: %s\n" % traceback.format_exc())
self.login_params["Email"] = username
self.login_params["Passwd"] = password
params = urllib.urlencode(self.login_params)
self.opener.open(self.url_authenticate, params)
def get_csv(self, throttle=False, **kwargs):
'''
Download CSV reports
'''
# Randomized download delay
if throttle:
r = random.uniform(0.5 * self.download_delay, 1.5 * self.download_delay)
time.sleep(r)
params = {
'export': 1
}
params.update(kwargs)
params = urllib.urlencode(params)
r = self.opener.open(self.url_download + params)
# Make sure everything is working ;)
if not r.info().has_key('Content-Disposition'):
print "You've exceeded your quota. Continue tomorrow..."
sys.exit(0)
if r.info().get('Content-Encoding') == 'gzip':
buf = StringIO( r.read())
f = gzip.GzipFile(fileobj=buf)
data = f.read()
else:
data = r.read()
myFile = open('trends_%s.csv' % '_'.join(['%s-%s' % (key, value) for (key, value) in kwargs.items()]), 'w')
myFile.write(data)
myFile.close()
Although I don't know python, I may have a solution. I am currently doing the same thing in C# and though I didn't get the .csv file, I got created a custom URL through code and then downloaded that HTML and saved to a text file (also through code). In this HTML (at line 12) is all the information needed to create the graph that is used on Google Trends. However, this has alot of unnecessary text within it that needs to be cut down. But either way, you end up with the same result. The Google Trends data. I posted a more detailed answer to my question here:
Downloading .csv file from Google Trends
There is an alternative module named pytrends - https://pypi.org/project/pytrends/ It is really cool. I would recommend this.
Example usage:
import numpy as np
import pandas as pd
from pytrends.request import TrendReq
pytrend = TrendReq()
#It is the term that you want to search
pytrend.build_payload(kw_list=["Eminem is the Rap God"])
# Find which region has searched the term
df = pytrend.interest_by_region()
df.to_csv("path\Eminem_InterestbyRegion.csv")
Potentially if you have a list of terms to search you could make use of "for loop" to automate the insights as per your wish.
In my django app, I have a view which accomplishes file upload.The core snippet is like this
...
if (request.method == 'POST'):
if request.FILES.has_key('file'):
file = request.FILES['file']
with open(settings.destfolder+'/%s' % file.name, 'wb+') as dest:
for chunk in file.chunks():
dest.write(chunk)
I would like to unit test the view.I am planning to test the happy path as well as the fail path..ie,the case where the request.FILES has no key 'file' , case where request.FILES['file'] has None..
How do I set up the post data for the happy path?Can somebody tell me?
I used to do the same with open('some_file.txt') as fp: but then I needed images, videos and other real files in the repo and also I was testing a part of a Django core component that is well tested, so currently this is what I have been doing:
from django.core.files.uploadedfile import SimpleUploadedFile
def test_upload_video(self):
video = SimpleUploadedFile("file.mp4", "file_content", content_type="video/mp4")
self.client.post(reverse('app:some_view'), {'video': video})
# some important assertions ...
In Python 3.5+ you need to use bytes object instead of str. Change "file_content" to b"file_content"
It's been working fine, SimpleUploadedFile creates an InMemoryFile that behaves like a regular upload and you can pick the name, content and content type.
From Django docs on Client.post:
Submitting files is a special case. To POST a file, you need only
provide the file field name as a key, and a file handle to the file
you wish to upload as a value. For example:
c = Client()
with open('wishlist.doc') as fp:
c.post('/customers/wishes/', {'name': 'fred', 'attachment': fp})
I recommend you to take a look at Django RequestFactory. It's the best way to mock data provided in the request.
Said that, I found several flaws in your code.
"unit" testing means to test just one "unit" of functionality. So,
if you want to test that view you'd be testing the view, and the file
system, ergo, not really unit test. To make this point more clear. If
you run that test, and the view works fine, but you don't have
permissions to save that file, your test would fail because of that.
Other important thing is test speed. If you're doing something like
TDD the speed of execution of your tests is really important.
Accessing any I/O is not a good idea.
So, I recommend you to refactor your view to use a function like:
def upload_file_to_location(request, location=None): # Can use the default configured
And do some mocking on that. You can use Python Mock.
PS: You could also use Django Test Client But that would mean that you're adding another thing more to test, because that client make use of Sessions, middlewares, etc. Nothing similar to Unit Testing.
I do something like this for my own event related application but you should have more than enough code to get on with your own use case
import tempfile, csv, os
class UploadPaperTest(TestCase):
def generate_file(self):
try:
myfile = open('test.csv', 'wb')
wr = csv.writer(myfile)
wr.writerow(('Paper ID','Paper Title', 'Authors'))
wr.writerow(('1','Title1', 'Author1'))
wr.writerow(('2','Title2', 'Author2'))
wr.writerow(('3','Title3', 'Author3'))
finally:
myfile.close()
return myfile
def setUp(self):
self.user = create_fuser()
self.profile = ProfileFactory(user=self.user)
self.event = EventFactory()
self.client = Client()
self.module = ModuleFactory()
self.event_module = EventModule.objects.get_or_create(event=self.event,
module=self.module)[0]
add_to_admin(self.event, self.user)
def test_paper_upload(self):
response = self.client.login(username=self.user.email, password='foz')
self.assertTrue(response)
myfile = self.generate_file()
file_path = myfile.name
f = open(file_path, "r")
url = reverse('registration_upload_papers', args=[self.event.slug])
# post wrong data type
post_data = {'uploaded_file': i}
response = self.client.post(url, post_data)
self.assertContains(response, 'File type is not supported.')
post_data['uploaded_file'] = f
response = self.client.post(url, post_data)
import_file = SubmissionImportFile.objects.all()[0]
self.assertEqual(SubmissionImportFile.objects.all().count(), 1)
#self.assertEqual(import_file.uploaded_file.name, 'files/registration/{0}'.format(file_path))
os.remove(myfile.name)
file_path = import_file.uploaded_file.path
os.remove(file_path)
I did something like that :
from django.core.files.uploadedfile import SimpleUploadedFile
from django.test import TestCase
from django.core.urlresolvers import reverse
from django.core.files import File
from django.utils.six import BytesIO
from .forms import UploadImageForm
from PIL import Image
from io import StringIO
def create_image(storage, filename, size=(100, 100), image_mode='RGB', image_format='PNG'):
"""
Generate a test image, returning the filename that it was saved as.
If ``storage`` is ``None``, the BytesIO containing the image data
will be passed instead.
"""
data = BytesIO()
Image.new(image_mode, size).save(data, image_format)
data.seek(0)
if not storage:
return data
image_file = ContentFile(data.read())
return storage.save(filename, image_file)
class UploadImageTests(TestCase):
def setUp(self):
super(UploadImageTests, self).setUp()
def test_valid_form(self):
'''
valid post data should redirect
The expected behavior is to show the image
'''
url = reverse('image')
avatar = create_image(None, 'avatar.png')
avatar_file = SimpleUploadedFile('front.png', avatar.getvalue())
data = {'image': avatar_file}
response = self.client.post(url, data, follow=True)
image_src = response.context.get('image_src')
self.assertEquals(response.status_code, 200)
self.assertTrue(image_src)
self.assertTemplateUsed('content_upload/result_image.html')
create_image function will create image so you don't need to give static path of image.
Note : You can update code as per you code.
This code for Python 3.6.
from rest_framework.test import force_authenticate
from rest_framework.test import APIRequestFactory
factory = APIRequestFactory()
user = User.objects.get(username='#####')
view = <your_view_name>.as_view()
with open('<file_name>.pdf', 'rb') as fp:
request=factory.post('<url_path>',{'file_name':fp})
force_authenticate(request, user)
response = view(request)
As mentioned in Django's official documentation:
Submitting files is a special case. To POST a file, you need only provide the file field name as a key, and a file handle to the file you wish to upload as a value. For example:
c = Client()
with open('wishlist.doc') as fp:
c.post('/customers/wishes/', {'name': 'fred', 'attachment': fp})
More Information: How to check if the file is passed as an argument to some function?
While testing, sometimes we want to make sure that the file is passed as an argument to some function.
e.g.
...
class AnyView(CreateView):
...
def post(self, request, *args, **kwargs):
attachment = request.FILES['attachment']
# pass the file as an argument
my_function(attachment)
...
In tests, use Python's mock something like this:
# Mock 'my_function' and then check the following:
response = do_a_post_request()
self.assertEqual(mock_my_function.call_count, 1)
self.assertEqual(
mock_my_function.call_args,
call(response.wsgi_request.FILES['attachment']),
)
if you want to add other data with file upload then follow the below method
file = open('path/to/file.txt', 'r', encoding='utf-8')
data = {
'file_name_to_receive_on_backend': file,
'param1': 1,
'param2': 2,
.
.
}
response = self.client.post("/url/to/view", data, format='multipart')`
The only file_name_to_receive_on_backend will be received as a file other params received normally as post paramas.
In Django 1.7 there's an issue with the TestCase wich can be resolved by using open(filepath, 'rb') but when using the test client we have no control over it. I think it's probably best to ensure file.read() returns always bytes.
source: https://code.djangoproject.com/ticket/23912, by KevinEtienne
Without rb option, a TypeError is raised:
TypeError: sequence item 4: expected bytes, bytearray, or an object with the buffer interface, str found
from django.test import Client
from requests import Response
client = Client()
with open(template_path, 'rb') as f:
file = SimpleUploadedFile('Name of the django file', f.read())
response: Response = client.post(url, format='multipart', data={'file': file})
Hope this helps.
Very handy solution with mock
from django.test import TestCase, override_settings
#use your own client request factory
from my_framework.test import APIClient
from django.core.files import File
import tempfile
from pathlib import Path
import mock
image_mock = mock.MagicMock(spec=File)
image_mock.name = 'image.png' # or smt else
class MyTest(TestCase):
# I assume we want to put this file in storage
# so to avoid putting garbage in our MEDIA_ROOT
# we're using temporary storage for test purposes
#override_settings(MEDIA_ROOT=Path(tempfile.gettempdir()))
def test_send_file(self):
client = APIClient()
client.post(
'/endpoint/'
{'file':image_mock},
format="multipart"
)
I am using Python==3.8.2 , Django==3.0.4, djangorestframework==3.11.0
I tried self.client.post but got a Resolver404 exception.
Following worked for me:
import requests
upload_url='www.some.com/oaisjdoasjd' # your url to upload
with open('/home/xyz/video1.webm', 'rb') as video_file:
# if it was a text file we would perhaps do
# file = video_file.read()
response_upload = requests.put(
upload_url,
data=video_file,
headers={'content-type': 'video/webm'}
)
I am using django rest framework and I had to test the upload of multiple files.
I finally get it by using format="multipart" in my APIClient.post request.
from rest_framework.test import APIClient
...
self.client = APIClient()
with open('./photo.jpg', 'rb') as fp:
resp = self.client.post('/upload/',
{'images': [fp]},
format="multipart")
I am using GraphQL, upload for test:
with open('test.jpg', 'rb') as fp:
response = self.client.execute(query, variables, data={'image': [fp]})
code in class mutation
#classmethod
def mutate(cls, root, info, **kwargs):
if image := info.context.FILES.get("image", None):
kwargs["image"] = image
TestingMainModel.objects.get_or_create(
id=kwargs["id"],
defaults=kwargs
)