使用 pycurl 为 url 传递多个参数

发布于 2024-10-04 01:48:04 字数 584 浏览 12 评论 0原文

我想使用具有多个参数的 URL 进行curl 调用。我在下面列出了代码。例如，是否有“curl -d @filter”的等效选项，或者我是否对参数进行 URL 编码？

SER = foobar
PASS = XXX

STREAM_URL = "http://status.dummy.com/status.json?userId=12&page=1"
class Client:   
          def __init__(self):
        self.buffer = ""
        self.conn = pycurl.Curl()
        self.conn.setopt(pycurl.USERPWD, "%s:%s" % (USER,PASS))
        self.conn.setopt(pycurl.URL, STREAM_URL)
        self.conn.setopt(pycurl.WRITEFUNCTION, self.on_receive)
        self.conn.perform()

    def on_receive(self,data):
        self.buffer += data

原文

I want to make a curl call with a URL that has multiple parameters. I have listed the code below. Is there an equivalent option for "curl -d @filter" for instance or do I have URL encode the parameters?

SER = foobar
PASS = XXX

STREAM_URL = "http://status.dummy.com/status.json?userId=12&page=1"
class Client:   
          def __init__(self):
        self.buffer = ""
        self.conn = pycurl.Curl()
        self.conn.setopt(pycurl.USERPWD, "%s:%s" % (USER,PASS))
        self.conn.setopt(pycurl.URL, STREAM_URL)
        self.conn.setopt(pycurl.WRITEFUNCTION, self.on_receive)
        self.conn.perform()

    def on_receive(self,data):
        self.buffer += data

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梦罢 2024-10-11 01:48:04

Pycurl 是 libcurl 的一个非常薄的包装。如果你能用 libcurl 做到这一点，你也可以用 pycurl 做到这一点。（大部分。）

例如：

pycurl.setopt对应libcurl中的curl_easy_setopt，其中option是用libcurl中的CURLOPT_*常量指定的，只不过CURLOPT_前缀已被删除。

请参阅： http://pycurl.sourceforge.net/doc/curlobject.html

那是也就是说，curl -d 选项用于发送 HTTP POST 请求...而不是您的示例显示的 GET 样式。

libcurl 确实期望它接收的 url 已经经过 URL 编码。如果需要，只需使用 http://docs.python.org/library/urllib.html 。

您问题中的示例 URL 已经有 2 个参数（userId 和 page）。

一般而言，格式为：URL 后跟“问号”，后跟由 & 符号连接的名称=值对。如果名称或值包含特殊字符，则需要对它们进行百分比编码。

只需使用 urlencode 函数：

>>> import urllib
>>> params = [('name1','value1'), ('name2','value2 with spaces & stuff')]
>>> pairs = urllib.urlencode(params)
>>> fullurl = 'http://status.dummy.com/status.json' + '?' + pairs
>>> print fullurl
http://status.dummy.com/status.json?name1=value1&name2=value2+with+spaces+%26+stuff
>>>

另请参阅 urllib.urlopen 函数。也许你根本不需要curl？（但我不知道您的申请...）

希望这有帮助。如果是这样，请标记已回答并告诉我。 :-)

Pycurl is a pretty thin wrapper for libcurl. If you can do it with libcurl, you can do it with pycurl. (Mostly.)

For instance:

pycurl.setopt corresponds to curl_easy_setopt in libcurl, where option is specified with the CURLOPT_* constants in libcurl, except that the CURLOPT_ prefix has been removed.

See: http://pycurl.sourceforge.net/doc/curlobject.html

That being said, the curl -d option is for sending HTTP POST requests... not the GET style your example shows.

libcurl does expect that urls it recives already be URL encoded. Just use http://docs.python.org/library/urllib.html if needed.

The sample URL in your question already has 2 parameters (userId and page).

In general the format is: URL followed by a 'question mark', followed by name=value pairs joined by an ampersand symbol. If the name or value contain special chars, you will need to percent-encoded them.

Just use the urlencode function:

>>> import urllib
>>> params = [('name1','value1'), ('name2','value2 with spaces & stuff')]
>>> pairs = urllib.urlencode(params)
>>> fullurl = 'http://status.dummy.com/status.json' + '?' + pairs
>>> print fullurl
http://status.dummy.com/status.json?name1=value1&name2=value2+with+spaces+%26+stuff
>>>

Also, see the urllib.urlopen function. Perhaps you do not need curl at all? (But I do not know your application...)

Hope this helps. If so, mark answered and let me know. :-)

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