HTTPS POST请求Python,返回.csv
问题描述:
我想向一个HTTPS站点发出一个POST请求,该站点应该以.csv文件响应。我有这样的Python代码:HTTPS POST请求Python,返回.csv
try:
#conn = httplib.HTTPSConnection(host="www.site.com", port=443)
=>给出一个BadStatusLine:'错误
conn = httplib.HTTPConnection("www.site.com");
params = urllib.urlencode({'val1':'123','val2':'abc','val3':'1b3'})
conn.request("POST", "/nps/servlet/exportdatadownload", params)
content = conn.getresponse()
print content.reason, content.status
print content.read()
conn.close()
except:
import sys
print sys.exc_info()[:2]
输出:
Found 302
<!DOCTYPE HTML PUBLIC "-//IETF//DTD HTML 2.0//EN">
<HTML><HEAD>
<TITLE>302 Found</TITLE>
</HEAD><BODY>
<H1>Found</H1>
The document has moved <A HREF="https://www.site.com/nps/servlet/exportdatadownload">here</A>.<P>
<HR>
<ADDRESS>Oracle-Application-Server-10g/10.1.3.5.0 Oracle-HTTP-Server Server at mp-www1.mrco.be Port 7778</ADDRESS>
</BODY></HTML>
我在做什么错?我如何“抓住”返回的.csv文件? 我尝试了Chrome浏览器扩展(高级休息客户端,这是工作...)的POST请求
答
带有302错误的HTML输出是因为您使用HTTP连接到网站而不是HTTPS,这是在你的代码在这里:
conn = httplib.HTTPConnection("www.site.com");
想必你这样做,因为你的代码的注释部分其他一些错误的。
如果我试图做到这一点,我会使用Requests,它带有非常明确的documentation。随着请求的代码会是这样的:
import requests
url = "https://www.example.com/nps/servlet/exportdatadownload"
payload = { "val1": "123", "val2": "abc", "val3": "1b3" }
r = requests.post(url, data=payload, verify=True)
CSV文件应在r.content,一个可以写入到文件中。
正如错误所示,Web服务器希望您通过https访问url - 当您尝试建立HTTPSConnection时,确切的错误是什么? – 2013-02-24 23:25:48
在命令行上,以下命令给你什么? 'curl -XPOST'https://www.site.com/nps/servlet/exportdatadownload?val1 = 123&val2 = abc&val3 = 1b3'' – Trevor 2013-02-24 23:25:53
@ Ali-AkberSaifee:'(,BadStatusLine(“ ''“,))' –
francisMi
2013-02-25 11:40:12