how to send a POST request in python equivalent with this curl command
curl -u "YOUR_USERNAME:YOUR_ACCESS_KEY" \
-X POST "https://api-cloud.browserstack.com/app-automate/upload" \
-F "url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk"
I tried code below:
resp=requests.post(URL,headers=
{'YOUR_USERNAME:YOUR_ACCESS_KEY'},
data=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk")
and its not working.
I don't know how to send this line "url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk"
in POST request. "url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk" this is the public url of apk. and want to upload at this url "https://api-cloud.browserstack.com/app-automate/upload"
And after applying answer from below, I find the solution Thank you Everyone. Answer is -
import urllib.request
#file will download in current working directory with name app-release.apk urllib.request.urlretrieve('https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk', 'app-release.apk')
test_file = open("app-release.apk", "rb")
URL = 'https://api-cloud.browserstack.com/app-automate/upload' response = requests.post(URL, files={'file': test_file, }, auth=('YOUR_USERNAME', 'YOUR_ACCESS_KEY'))
Use requests
You can do like this:
import requests
files = { 'file': ('url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk',
open('url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk',
'rb')),}
URL = 'https://api-cloud.browserstack.com/app-automate/upload'
response = requests.post(URL, files=files, auth=('YOUR_USERNAME', 'YOUR_ACCESS_KEY'))
print (response.text)
It should work fine now.
try this
import requests
files = {
'url': (None, 'https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk'),
}
response = requests.post('https://api-cloud.browserstack.com/app-automate/upload', files=files, auth=('YOUR_USERNAME', 'YOUR_ACCESS_KEY'))
or this
import requests
files = {
'url': 'https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk'
}
response = requests.post('https://api-cloud.browserstack.com/app-automate/upload', files=files, auth=('YOUR_USERNAME', 'YOUR_ACCESS_KEY'))
'''
And The answer of the solution, That is work for me is
import urllib.request
#file will download in current working directory with name app-release.apk
urllib.request.urlretrieve('https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk', 'app-release.apk')
test_file = open("app-release.apk", "rb")
URL = 'https://api-cloud.browserstack.com/app-automate/upload' response = requests.post(URL, files={'file': test_file, }, auth=('YOUR_USERNAME', 'YOUR_ACCESS_KEY'))
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