使用pycurl上傳文件

 

import pycurl
url = "http://www.xxxxx.com/upload.php"

field = "uploadFile"

file = "/home/rare/tmp/a.py"

c = pycurl.Curl()
c.setopt(c.POST, 1)
c.setopt(c.URL, url)
c.setopt(c.HTTPPOST, [(field, (c.FORM_FILE,  file))])
#c.setopt(c.VERBOSE, 1)
c.perform()
c.close()

其中url和field分別替換爲提交表單中action對應的腳本地址和file類型的input標籤的name值

file是需要上傳的文件路徑，需要注意在windows下反斜槓必須轉移，例如"c://a//b//c.txt"