I am using below code where I am using PUT api from POSTMAN to send a file to a machine hosting the api using python script
#app.route('/uploadFIle', methods=['PUT'])
def uploadFile():
chunk_size = 4096
with open("/Users/xyz/Documents/filename", 'wb') as f:
while True:
chunk = request.stream.read(chunk_size)
if len(chunk) == 0:
break
f.write(chunk)
return jsonify({"success":"File transfer initiated"})
Is there a way to get the original filename so that I can use the same while saving the file ?
Can do as below by passing name from PUT api itself, but is it the best solution ?
#app.route('/uploadFIle/<string:filename>', methods=['PUT'])
def uploadFile(filename):
Below is how I achieved it using flask -
Choose form-data under body in POSTMAN
You can give any key, i used 'file' as key, then choose option 'file' from drop down arrow in key column
Attach file under 'value' column and use below code to get the file name -
from flask import request
file = request.files['file']
file_name = file.filename
Related
I am posting file to python flask and then read its content using following code:
def post(self):
if 'file' not in request.files:
return {'error': 'no file'}
try:
f=open("text2.txt")
local_content=f.read()
content=request.files['file'].read().decode('utf-8')
if hash(content) != hash(local_content) :
return {'error': 'content changed','local':hash(local_content),'uploaded':hash(content)}
else:
return {'error': 'same','local':hash(local_content),'uploaded':hash(content)}
I also put same file text2.txt on server and read it locally using
local_content=f.read()
but both results are different.I tried comparing two string using following
if content != local_content
above conditions is always returning true.
but when I print both strings they are exactly same.
I am doing some processing on those strings and trying and both content and local_content produces different results.
So can anyone tell me why uploaded content is behaving differently than local content
Even I face this issue while uploading images using flask. The binary streams were different as compared to reading local file.
I solve this issue with file.seek(0)
files = request.files.getlist('files[]')
for file in files:
file.seek(0)
fileBytes = file.read()
image_stringRead = base64.b64encode(fileBytes)
I am trying to create a flask app that can be used to upload any user selected file to my azure storage. For some reason, the mime-type of the uploaded file is always set to 'application/octet-stream'. If I directly upload the file to azure using its UI, then the mime-type is correct. To solve this problem, I am trying to manually calculate the mimetype of the file and pass it as metadata.
The issue I am having is that I am not able to figure out a way to get the absolute filepath of the user selected file to be uploaded.
What I am looking for is the absolute path: path/to/file/doctest2.txt
Here is how the flask app looks like:
#app.route('/', methods=['GET', 'POST'])
def upload_file():
if request.method == 'POST':
file = request.files['file']
filename = secure_filename(file.filename)
fileextension = filename.rsplit('.',1)[1]
Randomfilename = id_generator()
filename = Randomfilename + '.' + fileextension
try:
blob_service.create_blob_from_stream(container, filename, file)
except Exception:
print 'Exception=' + Exception
pass
ref = 'http://'+ account + '.blob.core.windows.net/' + container + '/' + filename
seems like we can get the filename using f.filename, but I am not sure how to get the full path here.
Complete code can be found here:
https://github.com/codesagar/Azure-Blobs/blob/master/blob.py
The ultimate goal is to calculate the mimetype of the file to be uploaded.
I do have the file-blob(variable f). IS there a better way to get the mime from blob rather than hunting for the absolute file-path?
I solved my problem by using the following line of code:
mime_type = f.content_type
This gives me the mimetype of the file and eliminates the need for getting the file's absolute path.
def a():
import json
path=open('C:\\Users\\Bishal\\code\\57.json').read()
config=json.load(path)
for key in config:
return key
You have already read the file path=open('C:\Users\Bishal\code\57.json').read(), so when you try to load with json.load(path), the file pointer is at the end of the file; hence nothing gets loaded or parsed.
Either load the file directly into json, or read the contents and then parse the string with json.loads (note the s)
Option 1:
path = open(r'C:\Users\Bishal\code\57.json').read()
config = json.loads(path)
Option 2:
path = open(r'C:\Users\Bishal\code\57.json')
config = json.load(path)
path.close()
Then you can do whatever you like with the result:
for key,item in config.items():
print('{} - {}'.format(key, item))
A previous question asks how to retrieve at attachment from couchdb and display it in a flask application.
This question asks how to perform the opposite, i.e. how can an image be uploaded using flask and saved as a couchdb attachment.
Take a look at the example from WTF:
from werkzeug.utils import secure_filename
from flask_wtf.file import FileField
class PhotoForm(FlaskForm):
photo = FileField('Your photo')
#app.route('/upload/', methods=('GET', 'POST'))
def upload():
form = PhotoForm()
if form.validate_on_submit():
filename = secure_filename(form.photo.data.filename)
form.photo.data.save('uploads/' + filename)
else:
filename = None
return render_template('upload.html', form=form, filename=filename)
Take a look at the FileField api docs. There you have a stream method giving you access to the uploaded data. Instead of using the save method as in the example you can access the bytes from the stream, base64 encode it and save as an attachment in couchdb, e.g. Using put_attachment. Alternatively, the FileStorage api docs suggest you can use read() to retrieve the data.
Noob python user:
I've created file that extracts 10 tweets based on the api.search (not streaming api). I get a screen results, but cannot figure how to parse the output to save to csv. My error is TypeError: expected a character buffer object.
I have tried using .join(str(x) and get other errors.
My code is
import tweepy
import time
from tweepy import OAuthHandler
from tweepy import Cursor
#Consumer keys and access tokens, used for Twitter OAuth
consumer_key = ''
consumer_secret = ''
atoken = ''
asecret = ''
# The OAuth process that uses keys and tokens
auth = tweepy.OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(atoken, asecret)
# Creates instance to execute requests to Twitter API
api = tweepy.API(auth)
MarSec = tweepy.Cursor(api.search, q='maritime security').items(10)
for tweet in MarSec:
print " "
print tweet.created_at, tweet.text, tweet.lang
saveFile = open('MarSec.csv', 'a')
saveFile.write(tweet)
saveFile.write('\n')
saveFile.close()
Any help would be appreciated. I've gotten my Streaming API to work, but am having difficulty with this one.
Thanks.
tweet is not a string or a character buffer. It's an object. Replace your line with saveFile.write(tweet.text) and you'll be good to go.
saveFile = open('MarSec.csv', 'a')
for tweet in MarSec:
print " "
print tweet.created_at, tweet.text, tweet.lang
saveFile.write("%s %s %s\n"%(tweet.created_at, tweet.lang, tweet.text))
saveFile.close()
I just thought I'd put up another version for those who might want to save all
the attributes of a tweepy.models.Status object, if you're not yet sure which attributes of each tweet you want to save to file.
import json
search_results = []
for status in tweepy.Cursor(api.search, q=search_text).items(5000):
search_results.append(status._json)
with open('search_results.json', 'w') as f:
json.dump(search_results, f)
The first block will store the search results into a list of dictionaries, and the second block will output all the tweets into a json file.
Please beware, this might use up a lot of memory if the size of your search results is very big.
This is Twitter's classic error code when something is wrong while sending a wrong image.
Try to find images you are trying to upload and check the format of the images.
The only thing I did was erase the images that MY media player of Windows can´t read and that's all! the script run perfectly.