[英]how to collect text data from a URL link with or without ".html" in the link?
[英]Download text data from HTML link in Python
嗨,我想下载HTML链接上托管的带分隔符的文本。 (该链接只能在专用网络上访问,因此不能在此处共享)。
在R中,以下功能解决了这个问题(所有其他功能都给出了“未经授权的访问”或“ 401”错误)
url = 'https://dw-results.ansms.com/dw-platform/servlet/results? job_id=13802737&encoding=UTF8&mimeType=plain'
download.file(url, "~/insights_dashboard/testing_file.tsv")
a = read.csv("~/insights_dashboard/testing_file.tsv",header = T,stringsAsFactors = F,sep='\t')
我想在Python中做同样的事情,为此我使用了:
(A)urllib和request.get()
import urllib.request
url_get = requests.get(url, verify=False)
urllib.request.urlretrieve(url_get, 'C:\\Users\\cssaxena\\Desktop\\24.tsv')
(B)requests.get()和read.html
url='https://dw-results.ansms.com/dw-platform/servlet/results? job_id=13802737&encoding=UTF8&mimeType=plain'
s = requests.get(url, verify=False)
a = pd.read_html(io.StringIO(s.decode('utf-8')))
(C)使用wget:
import wget
url = 'https://dw-results.ansms.com/dw-platform/servlet/results? job_id=13802737&encoding=UTF8&mimeType=plain'
wget.download(url,--auth-no-challenge, 'C:\\Users\\cssaxena\\Desktop\\24.tsv')
要么
wget --server-response -owget.log "https://dw-results.ansms.com/dw-platform/servlet/results? job_id=13802737&encoding=UTF8&mimeType=plain"
注意:URL不需要任何凭据,浏览器可以访问该URL,并且可以使用R和download.file进行下载。 我正在寻找Python中的解决方案
def geturls(path):
yy=open(path,'rb').read()
yy="".join(str(yy))
yy=yy.split('<a')
out=[]
for d in yy:
z=d.find('href="')
if z>-1:
x=d[z+6:len(d)]
r=x.find('"')
x=x[:r]
x=x.strip(' ./')
#
if (len(x)>2) and (x.find(";")==-1):
out.append(x.strip(" /"))
out=set(out)
return(out)
pg="./test.html"# your html
url=geturls(pg)
print(url)
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