如何使用此 curl 命令在 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"
我試過下面的代碼：
resp=requests.post(URL,headers=
{'YOUR_USERNAME:YOUR_ACCESS_KEY'},
data=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk")
它不作業。
我不知道如何"url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk"在 POST 請求中發送這一行。“url=https://www.browserstack.com/app-automate/sample-apps/android/WikipediaSample.apk”這是apk的公共網址。并想在這個網址上傳“https://api-cloud.browserstack.com/app-automate/upload”
uj5u.com熱心網友回復：
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)
現在應該可以正常作業了。
uj5u.com熱心網友回復：
試試這個
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'))
'''
轉載請註明出處，本文鏈接：https://www.uj5u.com/ruanti/378706.html