I've been searching a lot and I haven't found an answer to what I'm looking for.
I'm trying to upload a file from /tmp to slack using python requests but I keep getting {"ok":false,"error":"no_file_data"} returned.
file={'file':('/tmp/myfile.pdf', open('/tmp/myfile.pdf', 'rb'), 'pdf')}
payload={
"filename":"myfile.pdf",
"token":token,
"channels":['#random'],
"media":file
}
r=requests.post("https://slack.com/api/files.upload", params=payload)
Mostly trying to follow the advice posted here
Sending files through http requires a bit more extra work than sending other data. You have to set content type and fetch the file and all that, so you can't just include it in the payload parameter in requests.
You have to give your file information to the files parameter of the .post method so that it can add all the file transfer information to the request.
my_file = {
'file' : ('/tmp/myfile.pdf', open('/tmp/myfile.pdf', 'rb'), 'pdf')
}
payload={
"filename":"myfile.pdf",
"token":token,
"channels":['#random'],
}
r = requests.post("https://slack.com/api/files.upload", params=payload, files=my_file)
Writing this post, to potentially save you all the time I've wasted. I did try to create a new file and upload it to Slack, without actually creating a file (just having it's content). Because of various and not on point errors from the Slack API I wasted few hours to find out that in the end, I had good code from the beginning and simply missed a bot in the channel.
This code can be used also to open an existing file, get it's content, modify and upload it to Slack.
Code:
from io import StringIO # this library will allow us to
# get a csv content, without actually creating a file.
sio = StringIO()
df.to_csv(sio) # save dataframe to CSV
csv_content = sio.getvalue()
filename = 'some_data.csv'
token=os.environ.get("SLACK_BOT_TOKEN")
url = "https://slack.com/api/files.upload"
request_data = {
'channels': 'C123456', # somehow required if you want to share the file
# it will still be uploaded to the Slack servers and you will get the link back
'content': csv_content, # required
'filename': filename, # required
'filetype': 'csv', # helpful :)
'initial_comment': comment, # optional
'text': 'File uploaded', # optional
'title': filename, # optional
#'token': token, # Don't bother - it won't work. Send a header instead (example below).
}
headers = {
'Authorization': f"Bearer {token}",
}
response = requests.post(
url, data=request_data, headers=headers
)
OFFTOPIC - about the docs
I just had a worst experience (probably of this year) with Slack's file.upload documentation. I think that might be useful for you in the future.
Things that were not working in the docs:
token - it cannot be a param of the post request, it must be a header. This was said in one of github bug reports by actual Slack employee.
channel_not_found - I did provide an existing, correct channel ID and got this message. This is somehow OK, because of security reasons (obfuscation), but why there is this error message then: not_in_channel - Authenticated user is not in the channel. After adding bot to the channel everything worked.
Lack of examples for using content param (that's why I am sharing my code with you.
Different codding resulted with different errors regarding form data and no info in the docs helped to understand what might be wrong, what encoding is required in which upload types.
The main issue is they do not version their API, change it and do not update docs, so many statements in the docs are false/outdated.
Base on the Slack API file.upload documentation
What you need to have are:
Token : Authentication token bearing required scopes.
Channel ID : Channel to upload the file
File : File to upload
Here is the sample code. I am using WebClient method in #slack/web-api package to upload it in slack channel.
import { createReadStream } from 'fs';
import { WebClient } from '#slack/web-api';
const token = 'token'
const channelId = 'channelID'
const web = new WebClient(token);
const uploadFileToSlack = async () => {
await web.files.upload({
filename: 'fileName',
file: createReadStream('path/file'),
channels: channelId,
});
}
Related
I want to upload files to Slack using Slack API and Python. I tried to use webhooks but it didn't help me send the file. I am able to send messages only but not files. Is there a way to achieve uploading the file?
Finally, this worked for me.
Steps
Create a Slack app. If you are not the admin you need to request the same.
Add the permissions to read, write.
Get the Authorization token from oauth & permissions
Use the following snippet to upload the file.
url = "https://slack.com/api/files.upload"
headers = {
"Authorization":"Bearer xxxx", ----> this is the token you receive from oauth & permissions. It generally starts with xox
}
payload = { "channels":"channelXYZ"}
file_upload = {
"file":("./hello-world.txt",
open("./hello-world.txt", 'rb'), 'text/plain')
}
response = requests.post(url, headers=headers, files=file_upload, data=payload)
I'm just working on API conection at my work. I already made some GET and PUT request, but now i have problem with POST. API documantation is here. And here is my code I test but get 400 bad request:
import requests
files = {'files': ('fv.pdf', open(r"C:\python\API\fv.pdf", 'rb'))}
data = {"order_documents":[{'file_name':"fv.pdf", 'type_code':'CUSTOMER_INVOICE' }]}
header = {
'Authorization': '###########################',
}
response = requests.post("https://######.com/api/orders/40100476277994-A/documents", headers=header, files = files, data = data)
print(response.status_code)
print(response.url)
Someone have any idea how i can handle with this?
Looks like you are missing the order_documents parameter, it needs to be an array and also needs to be called order_documents.
Try changing your data variable into:
data = {"order_documents": [ {'file_name':"fv.pdf", 'type_code':'CUSTOMER_INVOICE' } ] }
The API expects files as the parameter name and your dictionary sends file to the server. The parameter name files that you give to session.post is just for requests library and not the actual parameter sent to the server.
The API also expects multiple files in an array, so you need to change your files object.
files = [
('files', ('fv.pdf', open(r"C:\python\API\fv.pdf", 'rb')),
]
Also, I don't think you need to use requests.Session(), just use requests.post(), unless you're planning on using the session object multiple times for subsequent requests.
CKAN provides the ckanapi package for accessing the CKAN API via Python or the command line.
I can use it to download metadata, create resources, etc. But I can't create a package and upload resources to it in a single API call. (A package is also referred to as a dataset.)
Internally, ckanapi scans all keys moving any file-like parameters into a separate dict, which it passes to the requests.session.post(files=..) parameter.
This is the closest I can get but CKAN returns an HTTP 500 error (copied from this guide to requests):
with ckanapi.RemoteCKAN('http://myckan.example.com', apikey='real-key', user_agent=ua, username='joe', password='pwd') as ckan:
ckan.action.package_create(name='joe_data',
resources=('report.xls',
open('/path/to/file.xlsx', 'rb'),
'application/vnd.ms-excel',
{'Expires': '0'}))
I've also tried resources=open('path/file'), files=open('file'), shorter or longer tuples, but get the same 500 error.
The requests documentation says:
:param files: (optional) Dictionary of ``'filename': file-like-objects``
for multipart encoding upload.
I can't pass ckanapi resources={'filename': open('file')} as ckanapi doesn't detect the file, attempts to pass it to requests as a normal parameter, and fails ("BufferedReader is not JSON serializable" as it attempts to make the file a POST parameter). I get the same if I try to pass a list of files. But the API is able to create a package and add a number of resources in a single call.
So how do I create a package and multiple resources with a single ckanapi call?
I was curious about this and thought I'd put something together to test it. Unfortunately I haven't played with the CLI you mentioned. But I hope this will help you and others stumbling across this.
I am not positive but I'm guessing your resource dict isn't formatted properly. The resources needs to be a list of dictionaries.
Here's a ruby script to do the single api call insert (my preferred language at this time):
# Ruby script to create a package and resource in one api call.
# You can run this in https://repl.it/languages/ruby
# Don't forget to update URLs and API key.
require 'csv'
require 'json'
require 'net/http'
hash_to_json = {
"title" => 'test1',
"name" => 'test1',
"owner_org" => 'bbb9682e-b58c-4826-bf4b-b161581056be',
"resources" => [
{
"url" => 'http://www.resource_domain.com/doc.kml'
}
]
}.to_json
uri = URI('http://ckan_app_domain.com:5000/api/3/action/package_create')
Net::HTTP.start(uri.host, uri.port) do |http|
request = Net::HTTP::Post.new uri
request['Authorization'] = 'user-api-key'
request.body = hash_to_json
response = http.request request
puts response.body
end
And here's a plain python script to do the same thing (thank you CKAN docs for this template I modified)
#!/usr/bin/env python
import urllib2
import urllib
import json
import pprint
# Put the details of the dataset we're going to create into a dict.
dataset_dict = {
'name': 'my_dataset_name',
'notes': 'A long description of my dataset',
'owner_org': 'bbb9682e-b58c-4826-bf4b-b161581056be',
'resources': [
{
'url': 'example.com'
}
]
}
# Use the json module to dump the dictionary to a string for posting.
data_string = urllib.quote(json.dumps(dataset_dict))
# We'll use the package_create function to create a new dataset.
request = urllib2.Request(
'http://ckan_app_domain.com:5000/api/3/action/package_create')
# Creating a dataset requires an authorization header.
# Replace *** with your API key, from your user account on the CKAN site
# that you're creating the dataset on.
request.add_header('Authorization', 'user-api-key')
# Make the HTTP request.
response = urllib2.urlopen(request, data_string)
assert response.code == 200
# Use the json module to load CKAN's response into a dictionary.
response_dict = json.loads(response.read())
assert response_dict['success'] is True
# package_create returns the created package as its result.
created_package = response_dict['result']
pprint.pprint(created_package)
I am communicating with an API using HTTP.client in Python 3.6.2.
In order to upload a file it requires a three stage process.
I have managed to talk successfully using POST methods and the server returns data as I expect.
However, the stage that requires the actual file to be uploaded is a PUT method - and I cannot figure out how to syntax the code to include a pointer to the actual file on my storage - the file is an mp4 video file.
Here is a snippet of the code with my noob annotations :)
#define connection as HTTPS and define URL
uploadstep2 = http.client.HTTPSConnection("grabyo-prod.s3-accelerate.amazonaws.com")
#define headers
headers = {
'accept': "application/json",
'content-type': "application/x-www-form-urlencoded"
}
#define the structure of the request and send it.
#Here it is a PUT request to the unique URL as defined above with the correct file and headers.
uploadstep2.request("PUT", myUniqueUploadUrl, body="C:\Test.mp4", headers=headers)
#get the response from the server
uploadstep2response = uploadstep2.getresponse()
#read the data from the response and put to a usable variable
step2responsedata = uploadstep2response.read()
The response I am getting back at this stage is an
"Error 400 Bad Request - Could not obtain the file information."
I am certain this relates to the body="C:\Test.mp4" section of the code.
Can you please advise how I can correctly reference a file within the PUT method?
Thanks in advance
uploadstep2.request("PUT", myUniqueUploadUrl, body="C:\Test.mp4", headers=headers)
will put the actual string "C:\Test.mp4" in the body of your request, not the content of the file named "C:\Test.mp4" as you expect.
You need to open the file, read it's content then pass it as body. Or to stream it, but AFAIK http.client does not support that, and since your file seems to be a video, it is potentially huge and will use plenty of RAM for no good reason.
My suggestion would be to use requests, which is a way better lib to do this kind of things:
import requests
with open(r'C:\Test.mp4'), 'rb') as finput:
response = requests.put('https://grabyo-prod.s3-accelerate.amazonaws.com/youruploadpath', data=finput)
print(response.json())
I do not know if it is useful for you, but you can try to send a POST request with requests module :
import requests
url = ""
data = {'title':'metadata','timeDuration':120}
mp3_f = open('/path/your_file.mp3', 'rb')
files = {'messageFile': mp3_f}
req = requests.post(url, files=files, json=data)
print (req.status_code)
print (req.content)
Hope it helps .
I am trying to import a csv of responses into Qualtrics using the API shown here: https://api.qualtrics.com/docs/import-responses. But, since I'm a noob at Python and (by extension) at Requests, I'm having trouble figuring out why I keep getting a 413. I've gotten this far:
formTest = {
'surveyId': 'my_id',
'file': {
'value': open('dataFiles/myFile.csv', 'rb'),
'options': {
'contentType': 'text/csv'
}
}
}
headersTest = {
"X-API-TOKEN": "my_token",
'content-type': "multipart/form-data"
}
r = requests.request("POST", url, data=formTest, headers=headersTest)
print(r.text)
The format for the formTest variable is something I found when looking through other code bases for an angular implementation of this, which may not apply to a python version of the code. I can successfully use cUrl, but Python Requests, in my current situation is the way to go (for various reasons).
In a fit of desperation, I tried directly translating the cUrl request to python requests, but that didn't seem to help much either.
Has anyone done something like this before? I took a look at posts for importing contacts and the like, but there was no luck there either (since the data that needs to be sent is formatted differently). Is there something I am missing?
It's best not to mix post data and files but use two separate dictionaries. For the files you should use the files= parameter, because it encodes the POST data as a Multipart Form data and creates the required Content-Type headers.
import requests
url = 'Qualtrics API'
file_path = 'path/to/file'
file_name = 'file.name'
data = {'surveyId':'my_id'}
files = {'file' : (file_name, open(file_path, 'rb'), 'text/csv')}
headers = {'X-API-TOKEN': 'my_token'}
r = requests.post(url, data=data, files=files, headers=headers)
print(r.text)
The first value in files['file'] is the file name (optional), followed by the file object, followed by the file content type (optional).
You will find more info in the docs: Requests, POST a Multipart-Encoded File.