将curl查询转换为请求 [英] translate curl query to requests
问题描述
我正尝试在以下位置使用文档：
https：// pairbulkdata.uspto.gov/#/api-documentation
I was trying to get hand on using documentation at: https://pairbulkdata.uspto.gov/#/api-documentation
但是，当我尝试这些查询时，出现错误消息。
我正在尝试将curl查询转换为python请求。
However when I tried those query, I got error message. I am trying to translate curl query to python requests.
curl -i -X POST -H "Content-Type:application/json" -d '{"searchText":"medicine AND diabetes","qf": "patentTitle"}' http://pairbulkdata.uspto.gov/queries
这是我正在尝试的python代码：
Here is the python code that I am trying:
import requests
data = {"searchText":"medicine AND diabetes","qf": "patentTitle"}
url = "http://pairbulkdata.uspto.gov/queries"
header = {"Content-Type":"application/json"}
r = requests.post(url, data = data, headers=header)
但是我遇到了错误。
<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd">\n<HTML><HEAD><META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1">\n<TITLE>ERROR: The request could not be satisfied</TITLE>\n</HEAD><BODY>\n<H1>ERROR</H1>\n<H2>The request could not be satisfied.</H2>\n<HR noshade size="1px">\nBad request.\n<BR clear="all">\n<HR noshade size="1px">\n<PRE>\nGenerated by cloudfront (CloudFront)\nRequest ID: OIhwX7a3zVJq04M_qf0sjWhuke3fHb1-6wFJsN7UX_Rp2w_gzebTGA==\n</PRE>\n<ADDRESS>\n</ADDRESS>\n</BODY></HTML>
推荐答案
尝试使用可以将 curl 转换为请求 的转换器p>
http://curl.trillworks.com/
https：//shibukawa.github .io / curl_as_dsl / index.html
编辑：
我检查了文档，并且有指向使用API获取数据的页面的链接
I checked documentation and there is link to page which uses API to get data
https://pairbulkdata.uspto.gov/#/search?q=medicine ％20AND％20diabetes& sort = applId％20asc
似乎它在 / api / queries ，但文档显示 /查询
It seems it uses url with /api/queries but documentations shows /queries
-
此代码为我提供了一些JSON数据-所以可能可行
This code gives me some data as JSON - so probably it works
curl 'https://pairbulkdata.uspto.gov/api/queries' -X POST -H 'Content-Type: application/json' -d '{"searchText":"medicine AND diabetes","qf": "patentTitle"}'
-
这也给出了一些结果。
import requests
url = "https://pairbulkdata.uspto.gov/api/queries"
headers = {"Content-Type":"application/json"}
data = {"searchText":"medicine AND diabetes","qf": "patentTitle"}
r = requests.post(url, json=data, headers=headers)
print(r.text)
我使用
-
https：//而不是http：// -
/ api / queries而不是/ queries -
json =而不是data =
https://instead ofhttp:///api/queriesinstead of/queriesjson=instead ofdata=
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