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如何在 python Web 服务中使用 httplib 将带有文件对象的参数发布到 URL。

我正在使用以下脚本：

import httplib
import urllib
params = urllib.urlencode({"@str1":"string1", "@str2":"string2", "@file":"/local/path/to/file/in/client/machine", "@action":"action.php" })
headers = {"Content-type":"application/pdf , text/*" }
conn = httplib.HTTPConnection("192.168.2.17")
conn.request("POST", "/SomeName/action.php", params, headers)
response = conn.getresponse()
print response.status, response.reason
data = response.read()
data
conn.close()

我有以下输出：

200
OK
<html>.....some html code </html>

我写了一些 php 代码来保存这些字符串和数据库中的文件我的问题是，我只将文件路径作为一个刺痛而不是我的文件。可能我必须发送文件对象，例如，

file_obj = open("filename.txt","r")
conn.request("POST", "/SomeName/action.php", file_obj, headers)

但我想同时发送字符串和文件。有什么建议可以解决这个问题吗？

编辑 我将代码更改如下：当我直接使用 httplib 将 pdf 文件发送到我的服务器时，该文件保存为 BIN 文件。

def document_management_service(self,str_loc_file_path,*args):
    locfile = open(str_loc_file_path,'rb').read()
    host = "some.hostname.com"
    selector = "/api/?task=create"
    fields = [('docName','INVOICE'),('docNumber','DOC/3'),('cusName','name'),('cusNumber','C124'),('category','INVOICE'),('data','TIJO MRS,SOME/DATA/CONTENT,Blahblah,2584.00,blahblah'),('action','create')]
    files = [('strfile','File.pdf',locfile)]
    response = self.post_multipart(host, selector, fields, files)
    print response
    pass

def post_multipart(self,host, selector, fields, files):
    content_type, body = self.encode_multipart_formdata(fields, files)
    h = httplib.HTTP(host)
    h.set_debuglevel(1)
    h.putrequest('POST', selector)
    h.putheader('content-type', content_type)
    h.putheader('content-length', str(len(body)))
    h.putheader('Host', host)
    h.endheaders()
    h.send(body)
    errcode, errmsg, headers= h.getreply()
    return h.file.read()

def encode_multipart_formdata(self, fields, files):
    LIMIT = '----------lImIt_of_THE_fIle_eW_$'
    CRLF = '\r\n'
    L = []
    for (key, value) in fields:
        L.append('--' + LIMIT)
        L.append('Content-Disposition: form-data; name="%s"' % key)
        L.append('')
        L.append(value)
    for (key, filename, value) in files:
        L.append('--' + LIMIT)
        L.append('Content-Disposition: form-data; name="%s"; filename="%s"' % (key, filename))
        L.append('Content-Type: %s' % self.get_content_type(filename))
        L.append('')
        L.append(value)
    L.append('--' + LIMIT + '--')
    L.append('')
    body = CRLF.join(L)
    content_type = 'multipart/form-data; boundary=%s' % LIMIT
    return content_type, body

def get_content_type(self, filename):
    return mimetypes.guess_type(filename)[0] or 'application/octet-stream'

我已经调试了显示为的请求：

[('content-length', '4191'), ('accept-ranges', 'bytes'), ('server', 'Apache/2.2.12 (Ubuntu)'), ('last-modified', 'Tue, 23 Oct 2012 04:46:36 GMT'), ('etag', 'W/"567dd-105f-4ccb2a7a9a500"'), ('date', 'Tue, 23 Oct 2012 04:46:36 GMT'), ('content-type', 'application/pdf')]
multipart/form-data; boundary=----------lImIt_of_THE_fIle_eW_$

而且我没有尝试请求，因为我想用 httplib 解决这个问题（没有任何外部库）

4

2 回答 2

2

要在正文中发布参数和文件，您可以使用multipart/form-data内容类型：

#!/usr/bin/env python
import requests # $ pip install requests

file = 'file content as a file object or string'
r = requests.post('http://example.com/SomeName/action.php',
                  files={'file': ('filename.txt', file)},
                  data={'str1': 'string1', 'str2': 'string2'})
print(r.text) # response

requests.post向服务器发送如下内容：

POST /SomeName/action.php HTTP/1.1
Host: example.com
Content-Length: 449
Content-Type: multipart/form-data; boundary=f27f8ef67cac403aaaf433f83742bd64
Accept-Encoding: identity, deflate, compress, gzip
Accept: */*

--f27f8ef67cac403aaaf433f83742bd64
Content-Disposition: form-data; name="str2"
Content-Type: text/plain

string2
--f27f8ef67cac403aaaf433f83742bd64
Content-Disposition: form-data; name="str1"
Content-Type: text/plain

string1
--f27f8ef67cac403aaaf433f83742bd64
Content-Disposition: form-data; name="file"; filename="filename.txt"
Content-Type: text/plain

file content as a file object or string
--f27f8ef67cac403aaaf433f83742bd64--

要重现它，httplib请参阅POST form-data with Python example。

如果您的参数不包含太多数据，一个更简单的解决方案是将它们传递到 url 查询部分并让正文仅包含文件：

#!/usr/bin/env python
import urllib
import requests # $ pip install requests

params = {'str1': 'string1', 'str2': 'string2', 'filename': 'filename.txt'}
file = 'file content as a file object or string, etc'    
url = 'http://example.com/SomeName/action.php?' + urllib.urlencode(params)
r = requests.post(url, data=file, headers={'Content-Type': 'text/plain'})
print(r.text) # response

它对应于以下 HTTP 请求：

POST /SomeName/action.php?str2=string2&str1=string1&filename=filename.txt HTTP/1.1
Host: example.com
Content-Length: 39
Content-Type: text/plain
Accept-Encoding: identity, deflate, compress, gzip
Accept: */*

file content as a file object or string

httplib如果您需要，它应该更容易翻译。

于 2012-10-20T07:52:12.007 回答
0

以下代码还可以解决使用 httplib 传输文件与其他元数据的问题（没有任何外部库）：

def document_management_service_success(self,str_loc_file_path,*args):
    locfile = open(str_loc_file_path,'rb').read()
    str_loc_file = locfile.split('#end_pymotw_header')
    initial_data = str_loc_file[0]
    encoded_data = ''.join("{0:08b}".format(ord(c)) for c in initial_data)
    params = urllib.urlencode({'docName':'INVOICE', 'docNumber':'RF/2', 'cusName':'Tijo john', 'cusNumber':'C124', 'category':'INVOICE', 'data':encoded_data})
    headers = {"Accept": "Text/*","Content-Disposition":"form-data" ,"Content-Type": "application/x-www-form-urlencoded, application/pdf, form-data"}
    conn = httplib.HTTPConnection("efiling.nucoreindia.com")
    conn.connect()
    conn.set_debuglevel(1)
    conn.request("POST", "/api/?task=create", params, headers)
    response = conn.getresponse()
    print "Server Response status is"+str(response.status)+"and Reason is,"+str(response.reason)
    print response.getheaders()
    print response.read()
    pass
于 2012-10-24T08:39:19.710 回答