Question

当您点击＆＃34;我同意＆＃34;时，我试图通过模仿POST数据来绕过此页面http://casesearch.courts.state.md.us/casesearch/。我已经能够用PHP做到这一点但由于某种原因我无法在Python中重新创建请求。尽管发送了正确的POST数据，但响应始终是包含免责声明的页面。我想要登陆的页面的标题是＆＃34;马里兰州司法机构案例搜索标准＆＃34;。我使用的是Python 3。

    url = "http://casesearch.courts.state.md.us/casesearch/"
    postdata = parse.urlencode({'disclaimer':'Y','action':'Continue'})
    postdata = postdata.encode("utf-8")
    header = {"Content-Type":"application/x-www-form-urlencoded"}

    req = request.Request(url,data=postdata,headers=header)
    res = request.urlopen(req)
    print(res.read())

Answer 1

您希望发布到的网址实际上是http://casesearch.courts.state.md.us/casesearch/processDisclaimer.jis。试试这个：

url = "http://casesearch.courts.state.md.us/casesearch/processDisclaimer.jis"
postdata = parse.urlencode({'disclaimer':'Y','action':'Continue'})
postdata = postdata.encode("utf-8")
header = {"Content-Type":"application/x-www-form-urlencoded"}

req = request.Request(url,data=postdata,headers=header)
res = request.urlopen(req)
print(res.read())

HTTP请求不能正确模仿表单

1 个答案: