如何在 python 中為此 curl 命令發送發布請求

Question

如何在與此 curl 命令等效的 python 中發送 POST 請求

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"

我嘗試了下面的代碼：

resp=requests.post(URL,headers= 
    {'YOUR_USERNAME:YOUR_ACCESS_KEY'},
  data=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk") 

它不工作。

我不知道如何在 POST 請求中發送此行"url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk" 。 “url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk”這是apk的公開url。 並想在此 url “https://api-cloud.browserstack.com/app-automate/upload”上傳

從下面應用答案后，我找到了解決方案謝謝大家。 答案是——

導入 urllib.request

#file 將下載到當前工作目錄，名稱為 app-release.apk urllib.request.urlretrieve('https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk', 'app-發布.apk'）

test_file = open("app-release.apk", "rb")

URL = 'https://api-cloud.browserstack.com/app-automate/upload' 響應 = requests.post(URL, files={'file': test_file, }, auth=('YOUR_USERNAME', 'YOUR_ACCESS_KEY' ))

Answer 1

使用請求

你可以這樣做：

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)

它現在應該可以正常工作了。

Answer 2

嘗試這個

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'))

或這個

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'))
'''

Answer 3

解決方案的答案，這對我有用的是

導入 urllib.request

#file 將下載到當前工作目錄，名稱為 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' 響應 = requests.post(URL, files={'file': test_file, }, auth=('YOUR_USERNAME', 'YOUR_ACCESS_KEY' ))