files = {'file': ('output.mp3', open('output.mp3','rb'), 'audio/mpeg')}
I am using this for POST request but after usage trying to delete it with os.remove and it gives "it's used by a process". How to close file after?
You can use with ...:
import os
import requests
# open the file and send it
with open("output.mp3", "rb") as my_file:
files = {"file": ("output.mp3", my_file, "audio/mpeg")}
r = requests.post(url, files=files)
# ...
# file is closed now, remove it
os.remove("output.mp3")
Related
import glob
import os
import requests
import shutil
class file_service:
def file():
dir_name = '/Users/TEST/Downloads/TU'
os.chdir(dir_name)
pattern='TU_*.csv'
for x in glob.glob(pattern):
file_name=os.path.join(dir_name,x)
print (file_name)
from datetime import date
dir_name_backup = '/Users/Zill/Downloads/backup'
today = date.today()
backup_file_name = f'Backup_TU_{today.year}{today.month:02}{today.day:02}.csv'
backup_file_name_directory= os.path.join(dir_name_backup,backup_file_name)
print(backup_file_name_directory)
newPath = shutil.copy(file_name, backup_file_name_directory)
url = "google.com"
payload = {'name': 'file'}
files = [
('file', open(file_name,'rb'))
]
headers = {
'X-API-TOKEN': '12312'
}
response = requests.request("POST", url, headers=headers, data = payload, files = files)
print(response.text.encode('utf8'))
files.close()
os.remove(file_name)
file()
To provide an overall context, I am trying to retrieve a file from my OS and using POST method I am trying to post the content of the file into an application. Its working as expected so far, the details are getting pushed into application as expected. As part of my next step I am trying to remove the file from my existing directory using os.remove(). But I am getting a Win32 error as my file is not closed when it was opened in read-only mode in the POST call. I am trying to close it but I am unable to do so.
Can anyone please help me out with it.
Thanks!
I'm not sure I understand your code correctly. Could you try replacing
files.close()
with
for _, file in files:
file.close()
and check if it works?
Explanation:
In
files = [('file', open(file_name,'rb'))]
you create a list containing exactly one tuple that has the string 'file' as first element and a file object as second element:
[('file', file_object)]
The loop takes the tuple from the list, ignores its first element (_), takes its second element, the file object, and uses its close method to close it.
I've just now realised the list contains only one tuple. So there's no need for a loop:
files[0][1].close()
should do it.
The best way would be to use with (the file gets automatically closed once you leave the with block):
payload = {'name': 'file'}
with open(file_name, 'rb') as file:
files = [('file', file)]
headers = {'X-API-TOKEN': '12312'}
response = requests.request("POST", url, headers=headers, data = payload, files = files)
print(response.text.encode('utf8'))
os.remove(file_name)
I am trying to fetch the data of a pdf file available online
I have tried
import requests
response = requests.get("http://imdagrimet.gov.in/sites/default/files/daas_bulletin/District%20Advisory%20patna_17.pdf")
print(response.content)
but it gives a byte object as a response, and I am not able to decode that
You should write the data inside a file in order to be able to get it.
Like this:
with open('/District_Advisory_patna_17.pdf', 'wb') as f:
f.write(response.content)
Try to write your data to file:
import requests
import shutil
url = 'your url'
r = requests.get(url, stream=True)
if r.status_code == 200:
with open(file_path, 'wb') as f:
r.raw.decode_content = True
shutil.copyfileobj(r.raw, f)
I try to download img from url, add it to zip archive and then response this archive by Django HttpResponse.
import os
import requests
import zipfile
from django.http import HttpResponse
url = 'http://some.link/img.jpg'
file = requests.get(url)
data = file.content
rf = open('tmp/pic1.jpg', 'wb')
rf.write(data)
rf.close()
zipf = zipfile.ZipFile('tmp/album.zip', 'w') # This file is ok
filename = os.path.basename(os.path.normpath('tmp/pic1.jpg'))
zipf.write('tmp/pic1.jpg', filename)
zipf.close()
resp = HttpResponse(open('tmp/album.zip', 'rb'))
resp['Content-Disposition'] = 'attachment; filename=album.zip'
resp['Content-Type'] = 'application/zip'
return resp # Got corrupted zip file
When I save file to tmp folder - it's ok, I can extract it.
But when I response this file I get 'Error 1/2/21' on MacOS or Unexpected EOF if I try to open in Atom editor (just for test).
I also used StringIO instead of saving zip file, but it doesn't influence the result.
If you're using Python 3, you'd do it like this:
import os, io, zipfile, requests
from django.http import HttpResponse
# Get file
url = 'https://some.link/img.jpg'
response = requests.get(url)
# Get filename from url
filename = os.path.split(url)[1]
# Create zip
buffer = io.BytesIO()
zip_file = zipfile.ZipFile(buffer, 'w')
zip_file.writestr(filename, response.content)
zip_file.close()
# Return zip
response = HttpResponse(buffer.getvalue())
response['Content-Type'] = 'application/x-zip-compressed'
response['Content-Disposition'] = 'attachment; filename=album.zip'
return response
That's without saving the file. Downloaded file goes directly to io.
To response saved file, use this syntax:
response = HttpResponse(open('path/to/file', 'rb').read())
What I am trying to do is loop through a list of URL to download a series of .pdfs, and save them to a .zip. At the moment I am just trying to test code using just one URL. The ERROR I am getting is:
Traceback (most recent call last):
File "I:\test_pdf_download_zip.py", line 36, in <module>
zip_file(zipfile_name, url)
File "I:\test_pdf_download_zip.py", line 30, in zip_file
myzip.write(dowload_pdf(url))
TypeError: expected a string or other character buffer object
Would someone know how to pass .pdf request to the .zip correctly (avoiding the error above) in order for me to append it, or know if it is possible to do this?
import os
import zipfile
import requests
output = r"I:"
# File name of the zipfile
zipfile_name = os.path.join(output, "test.zip")
# Random test pdf
url = r"http://www.pdf995.com/samples/pdf.pdf"
def create_zipfile(zipfile_name):
zipfile.ZipFile(zipfile_name, "w")
def dowload_pdf(url):
response = requests.get(url, stream=True)
with open('test.pdf', 'wb') as f:
f.write(response.content)
def zip_file(zip_name, url):
with open(zip_name,'a') as myzip:
myzip.write(dowload_pdf(url))
if __name__ == "__main__":
create_zipfile(zipfile_name)
zip_file(zipfile_name, url)
print("Done")
Your download_pdf() function is saving a file but it doesn't return anything. You need to modify it so it actually returns the file path to myzip.write(). You don't want to hardcode test.pdf but pass unique paths to your download function so you don't end up with multiple test.pdf in your archive.
def dowload_pdf(url, path):
response = requests.get(url, stream=True)
with open(path, 'wb') as f:
f.write(response.content)
return path
I'm using Requests to upload a PDF to an API. It is stored as "response" below. I'm trying to write that out to Excel.
import requests
files = {'f': ('1.pdf', open('1.pdf', 'rb'))}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
file = open("out.xls", "w")
file.write(response)
file.close()
I'm getting the error:
file.write(response)
TypeError: expected a character buffer object
I believe all the existing answers contain the relevant information, but I would like to summarize.
The response object that is returned by requests get and post operations contains two useful attributes:
Response attributes
response.text - Contains str with the response text.
response.content - Contains bytes with the raw response content.
You should choose one or other of these attributes depending on the type of response you expect.
For text-based responses (html, json, yaml, etc) you would use response.text
For binary-based responses (jpg, png, zip, xls, etc) you would use response.content.
Writing response to file
When writing responses to file you need to use the open function with the appropriate file write mode.
For text responses you need to use "w" - plain write mode.
For binary responses you need to use "wb" - binary write mode.
Examples
Text request and save
# Request the HTML for this web page:
response = requests.get("https://stackoverflow.com/questions/31126596/saving-response-from-requests-to-file")
with open("response.txt", "w") as f:
f.write(response.text)
Binary request and save
# Request the profile picture of the OP:
response = requests.get("https://i.stack.imgur.com/iysmF.jpg?s=32&g=1")
with open("response.jpg", "wb") as f:
f.write(response.content)
Answering the original question
The original code should work by using wb and response.content:
import requests
files = {'f': ('1.pdf', open('1.pdf', 'rb'))}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
file = open("out.xls", "wb")
file.write(response.content)
file.close()
But I would go further and use the with context manager for open.
import requests
with open('1.pdf', 'rb') as file:
files = {'f': ('1.pdf', file)}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
with open("out.xls", "wb") as file:
file.write(response.content)
You can use the response.text to write to a file:
import requests
files = {'f': ('1.pdf', open('1.pdf', 'rb'))}
response = requests.post("https://pdftables.com/api?&format=xlsx-single",files=files)
response.raise_for_status() # ensure we notice bad responses
with open("resp_text.txt", "w") as file:
file.write(response.text)
As Peter already pointed out:
In [1]: import requests
In [2]: r = requests.get('https://api.github.com/events')
In [3]: type(r)
Out[3]: requests.models.Response
In [4]: type(r.content)
Out[4]: str
You may also want to check r.text.
Also: https://2.python-requests.org/en/latest/user/quickstart/