The error that I'm getting on Postman
My Code:
#app.route('/getcoordinates', methods=['GET', 'POST'])
def get_image():
if request.method == 'POST':
file = request.files['image'].read()
#use numpy to construct an array from the bytes
x = np.fromstring(file, dtype='uint8')
#decode the array into an image
img = cv2.imdecode(x, cv2.IMREAD_UNCHANGED)
#In output we get the x&y coordinates of the face bounding box
output = give_coordinates(img)
else:
return "Error: No image provided. Please specify a image."
return jsonify(output)
Or can anyone tell me how to give a picture in POST request because I think that's where it's going wrong.
Possible answered here: "Post Image data using POSTMAN"
That's not how you send file on postman. What you did is sending a string which is the path of your image, nothing more.
What you should do is;
After setting request method to POST, click to the 'body' tab.
Select form-data. At first line, you'll see text boxes named key and value. Write 'image' to the key. You'll see value type which is set to 'text' as default. Make it File and upload your file.
Then select 'raw' and paste your json file. Also just next to the binary choice, You'll see 'Text' is clicked. Make it JSON.
Related
I have a function to create a simple PDF. But when working on special characters, it returns something like that. How do I correctly save characters such as śćźż in my pdf file?
I tried to change the font type using setFont (Helvetica, TimesRoman) according this doc but I was not able to get the expected results.
Views.py (oficial doc)
def some_view_aa(request):
# Create a file-like buffer to receive PDF data.
buffer = io.BytesIO()
# Create the PDF object, using the buffer as its "file."
p = canvas.Canvas(buffer)
# Draw things on the PDF. Here's where the PDF generation happens.
# See the ReportLab documentation for the full list of functionality.
p.drawString(100, 100, "Hello AZX AĄĄŻĄ world.")
# Close the PDF object cleanly, and we're done.
p.showPage()
p.save()
# FileResponse sets the Content-Disposition header so that browsers
# present the option to save the file.
buffer.seek(0)
return FileResponse(buffer, as_attachment=True, filename='hello.pdf')
I'm saving the PIL image to a io.BytesIO() object.
imgByteArr = io.BytesIO()
img.save(imgByteArr, format=format)
Then trying to return the image to the user.
return send_file(img.getvalue(), mimetype="image/" + img_details["ext"].lower())
But I'm getting the error
TypeError: The view function did not return a valid response. The function either returned None or ended without a return statement.
I dont want to send as an attachment, i want the image to be displayed on the page.
Does any one know if it is possible without saving to disk first?
I was missing "seek"
imgByteArr = io.BytesIO()
img.save(imgByteArr, format=format)
imgByteArr.seek(0)
return send_file(imgByteArr, mimetype="image/" + img_details["ext"].lower())
This now works.
I'm writing a Django function that takes some user input, and generates a pdf for the user. However, the process for generating the pdf is quite intensive, and I'll get a lot of repeated requests so I'd like to store the generated pdfs on the server and check if they already exist before generating them.
The problem is that django-wkhtmltopdf (which I'm using for generation) is meant to return to the user directly, and I'm not sure how to store it on the file.
I have the following, which works for returning a pdf at /pdf:
urls.py
urlpatterns = [
url(r'^pdf$', views.createPDF.as_view(template_name='site/pdftemplate.html', filename='my_pdf.pdf'))
]
views.py
class createPDF(PDFTemplateView):
filename = 'my_pdf.pdf'
template_name = 'site/pdftemplate.html'
So that works fine to create a pdf. What I'd like is to call that view from another view and save the result. Here's what I've got so far:
#Create pdf
pdf = createPDF.as_view(template_name='site/pdftemplate.html', filename='my_pdf.pdf')
pdf = pdf(request).render()
pdfPath = os.path.join(settings.TEMP_DIR,'temp.pdf')
with open(pdfPath, 'w') as f:
f.write(pdf.content)
This creates temp.pdf and is about the size I'd expect but the file isn't valid (it renders as a single completely blank page).
Any suggestions?
Elaborating on the previous answer given: to generate a pdf file and save to disk do this anywhere in your view:
...
context = {...} # build your context
# generate response
response = PDFTemplateResponse(
request=self.request,
template=self.template_name,
filename='file.pdf',
context=context,
cmd_options={'load-error-handling': 'ignore'})
# write the rendered content to a file
with open("file.pdf", "wb") as f:
f.write(response.rendered_content)
...
I have used this code in a TemplateView class so request and template fields were set like that, you may have to set it to whatever is appropriate in your particular case.
Well, you need to take a look to the code of wkhtmltopdf, first you need to use the class PDFTemplateResponse in wkhtmltopdf.views to get access to the rendered_content property, this property get us access to the pdf file:
response = PDFTemplateResponse(
request=<your_view_request>,
template=<your_template_to_render>,
filename=<pdf_filename.pdf>,
context=<a_dcitionary_to_render>,
cmd_options={'load-error-handling': 'ignore'})
Now you could use the rendered_content property to get access to the pdf file:
mail.attach('pdf_filename.pdf', response.rendered_content, 'application/pdf')
In my case I'm using this pdf to attach to an email, you could store it.
I'm using the following code (from django management commands) to listen to the Twitter stream - I've used the same code on a seperate command to track keywords successfully - I've branched this out to use location, and (apparently rightly) wanted to test this out without disrupting my existing analysis that's running.
I've followed the docs and have made sure the box is in Long/Lat format (in fact, I'm using the example long/lat from the Twitter docs now). It looks broadly the same as the question here, and I tried using their version of the code from the answer - same error. If I switch back to using 'track=...', the same code works, so it's a problem with the location filter.
Adding a print debug inside streaming.py in tweepy so I can see what's happening, I print out the self.parameters self.url and self.headers from _run, and get:
{'track': 't,w,i,t,t,e,r', 'delimited': 'length', 'locations': '-121.7500,36.8000,-122.7500,37.8000'}
/1.1/statuses/filter.json?delimited=length and
{'Content-type': 'application/x-www-form-urlencoded'}
respectively - seems to me to be missing the search for location in some way shape or form. I don't believe I'm/I'm obviously not the only one using tweepy location search, so think it's more likely a problem in my use of it than a bug in tweepy (I'm on 2.3.0), but my implementation looks right afaict.
My stream handling code is here:
consumer_key = 'stuff'
consumer_secret = 'stuff'
access_token='stuff'
access_token_secret_var='stuff'
import tweepy
import json
# This is the listener, resposible for receiving data
class StdOutListener(tweepy.StreamListener):
def on_data(self, data):
# Twitter returns data in JSON format - we need to decode it first
decoded = json.loads(data)
#print type(decoded), decoded
# Also, we convert UTF-8 to ASCII ignoring all bad characters sent by users
try:
user, created = read_user(decoded)
print "DEBUG USER", user, created
if decoded['lang'] == 'en':
tweet, created = read_tweet(decoded, user)
print "DEBUG TWEET", tweet, created
else:
pass
except KeyError,e:
print "Error on Key", e
pass
except DataError, e:
print "DataError", e
pass
#print user, created
print ''
return True
def on_error(self, status):
print status
l = StdOutListener()
auth = tweepy.OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_token, access_token_secret_var)
stream = tweepy.Stream(auth, l)
#locations must be long, lat
stream.filter(locations=[-121.75,36.8,-122.75,37.8], track='twitter')
The issue here was the order of the coordinates.
Correct format is:
SouthWest Corner(Long, Lat), NorthEast Corner(Long, Lat). I had them transposed. :(
The streaming API doesn't allow to filter by location AND keyword simultaneously.
you must refer to this answer i had the same problem earlier
https://stackoverflow.com/a/22889470/4432830
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).