Question

我正在使用PyCurl以文件作为附件发送POST请求：

d = pycurl.Curl()
d.setopt(pycurl.URL, url)
# d.setopt(pycurl.RETURNTRANSFER, True)
d.setopt(pycurl.POST, True)
d.setopt(pycurl.POSTFIELDS, {filename: "@" + filename})
b = StringIO.StringIO()
d.setopt(pycurl.WRITEFUNCTION, b.write)
d.perform()
d.close()
message = b.getvalue()

我得到了：

Something went wrong, invalid arguments to setopt
Traceback (most recent call last):
  File "hasoff.py", line 214, in create_offers_for_advertiser
    if filename:
TypeError: invalid arguments to setopt

怎么了？

Answer 1

pycurl.POSTFIELDS需要网址编码数据。 From docs：

post_data = {'field': 'value'}
postfields = urlencode(post_data)
c.setopt(c.POSTFIELDS, postfields)

您无法使用pycurl.POSTFIELDS发送文件。

可以使用pycurl.HTTPPOST发送文件。 From docs：

c.setopt(c.HTTPPOST, [
    ('fileupload', (
        c.FORM_FILE, __file__,
        c.FORM_FILENAME, 'helloworld.py',
        c.FORM_CONTENTTYPE, 'application/x-python',
    )),
])

PyCurl TypeError：setopt的无效参数

1 个答案: