谁能告诉我如何将下面的CURL命令转换为Python3请求模块?
卷曲命令:
curl --header "Accept=application/json" --form
[email protected] http://189.167.1.103/print_file_uploads
成功:
import os
ip_address = 189.167.1.103
file_name = "D:\networking\a.gcode"
os.system("curl --header "Accept=application/json" --form
print_file=@%shttp://%s/print_file_uploads"%(file_name,ip_address))
但我不希望它与OS模块完成。你能帮我在python3请求中实现相同的功能吗?
首先,你可以阅读this
因此,请求的示例是:
import requests
response = requests.get('https://github.com/timeline.json')
response.json()
可能的解决方案是:
import requests
url = 'http://189.167.1.103/print_file_uploads'
payload = open("D:\networking\a.gcode")
headers = {'Accept: application/json'}
response = requests.post(url, data=payload, headers=headers)
这可能有效 -
导入请求
access_token ='xxxx'device_id ='xxxx'
address ='https://api.spark.io/v1/devices/ {0} / update /'。format(device_id)data = {'access_token':access_token,'args':'000000000000000'}
r = requests.post(地址,数据=数据)
打印(r.text)