pycurl：RETURNTRANSFER选项不存在

发布于 2024-07-20 05:34:15 字数 582 浏览 12 评论 0原文

我正在使用 pycurl 访问 JSON Web API，但是当我尝试使用以下命令

ocurl.setopt(pycurl.URL, gaurl)       # host + endpoint
ocurl.setopt(pycurl.RETURNTRANSFER, 1)
ocurl.setopt(pycurl.HTTPHEADER, gaheader) # Send extra headers
ocurl.setopt(pycurl.CUSTOMREQUEST, "POST") # HTTP POST req
ocurl.setopt(pycurl.CONNECTTIMEOUT, 2)

并执行脚本时，它失败了。

File "getdata.py", line 46, in apicall
ocurl.setopt(pycurl.RETURNTRANSFER, 1)
AttributeError: 'module' object has no attribute 'RETURNTRANSFER'

我不知道发生了什么，也不知道为什么 RETURNTRANSFER 似乎不存在，而所有其他选项都存在。

原文

I'm using pycurl to access a JSON web API, but when I try to use the following:

ocurl.setopt(pycurl.URL, gaurl)       # host + endpoint
ocurl.setopt(pycurl.RETURNTRANSFER, 1)
ocurl.setopt(pycurl.HTTPHEADER, gaheader) # Send extra headers
ocurl.setopt(pycurl.CUSTOMREQUEST, "POST") # HTTP POST req
ocurl.setopt(pycurl.CONNECTTIMEOUT, 2)

and execute the script, it fails.

File "getdata.py", line 46, in apicall
ocurl.setopt(pycurl.RETURNTRANSFER, 1)
AttributeError: 'module' object has no attribute 'RETURNTRANSFER'

I haven't a clue what's going on, and why RETURNTRANSFER doesn't appear to exist while all the other options do.

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再可℃爱ぅ一点好了 2024-07-27 05:34:15

该手册显示的用法是类似这样的：

>>> import pycurl
>>> import StringIO
>>> b = StringIO.StringIO()
>>> conn = pycurl.Curl()
>>> conn.setopt(pycurl.URL, 'http://www.example.org')
>>> conn.setopt(pycurl.WRITEFUNCTION, b.write)
>>> conn.perform()
>>> print b.getvalue()
<HTML>
<HEAD>
  <TITLE>Example Web Page</TITLE>
</HEAD>
<body>
<p>You have reached this web page by typing "example.com",
"example.net",
  or "example.org" into your web browser.</p>
<p>These domain names are reserved for use in documentation and are not availabl
e
  for registration. See <a href="http://www.rfc-editor.org/rfc/rfc2606.txt">RFC

  2606</a>, Section 3.</p>
</BODY>
</HTML>

似乎有点迂回，但我不是 PycURL 的忠实粉丝...

The manual shows the usage being something like this:

>>> import pycurl
>>> import StringIO
>>> b = StringIO.StringIO()
>>> conn = pycurl.Curl()
>>> conn.setopt(pycurl.URL, 'http://www.example.org')
>>> conn.setopt(pycurl.WRITEFUNCTION, b.write)
>>> conn.perform()
>>> print b.getvalue()
<HTML>
<HEAD>
  <TITLE>Example Web Page</TITLE>
</HEAD>
<body>
<p>You have reached this web page by typing "example.com",
"example.net",
  or "example.org" into your web browser.</p>
<p>These domain names are reserved for use in documentation and are not availabl
e
  for registration. See <a href="http://www.rfc-editor.org/rfc/rfc2606.txt">RFC

  2606</a>, Section 3.</p>
</BODY>
</HTML>

Seems a little roundabout, but I'm not a big fan of PycURL...

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