我有以下蜘蛛:
class Downloader(scrapy.Spider):
name = "sor_spider"
download_folder = FOLDER
def get_links(self):
df = pd.read_excel(LIST)
return df["Value"].loc
def start_requests(self):
urls = self.get_links()
for url in urls.iteritems():
index = {"index" : url[0]}
yield scrapy.Request(url=url[1], callback=self.download_file, errback=self.errback_httpbin, meta=index, dont_filter=True)
def download_file(self, response):
url = response.url
index = response.meta["index"]
content_type = response.headers['Content-Type']
download_path = os.path.join(self.download_folder, r"{}".format(str(index)))
with open(download_path, "wb") as f:
f.write(response.body)
yield LinkCheckerItem(index=response.meta["index"], url=url, code="downloaded")
def errback_httpbin(self, failure):
yield LinkCheckerItem(index=failure.request.meta["index"], url=failure.request.url, code="error")
它应该:
LIST
)阅读excel FOLDER
LinkCheckerItem
中记录结果(我正在将其导出到csv)通常可以正常工作,但是我的列表中包含不同类型的文件-zip,pdf,doc等。
这些是我的LIST
中的链接示例:
https://disclosure.1prime.ru/Portal/GetDocument.aspx?emId=7805019624&docId=2c5fb68702294531afd03041e877ca84
http://www.e-disclosure.ru/portal/FileLoad.ashx?Fileid=1173293
http://www.e-disclosure.ru/portal/FileLoad.ashx?Fileid=1263289
https://disclosure.1prime.ru/Portal/GetDocument.aspx?emId=7805019624&docId=eb9f06d2b837401eba9c66c8bf5be813
http://e-disclosure.ru/portal/FileLoad.ashx?Fileid=952317
http://e-disclosure.ru/portal/FileLoad.ashx?Fileid=1042224
https://www.e-disclosure.ru/portal/FileLoad.ashx?Fileid=1160005
https://www.e-disclosure.ru/portal/FileLoad.ashx?Fileid=925955
https://www.e-disclosure.ru/portal/FileLoad.ashx?Fileid=1166563
http://npoimpuls.ru/templates/npoimpuls/material/documents/%D0%A1%D0%BF%D0%B8%D1%81%D0%BE%D0%BA%20%D0%B0%D1%84%D1%84%D0%B8%D0%BB%D0%B8%D1%80%D0%BE%D0%B2%D0%B0%D0%BD%D0%BD%D1%8B%D1%85%20%D0%BB%D0%B8%D1%86%20%D0%BD%D0%B0%2030.06.2016.pdf
http://нпоимпульс.рф/assets/download/sal30.09.2017.pdf
http://www.e-disclosure.ru/portal/FileLoad.ashx?Fileid=1166287
我希望它以原始扩展名保存文件,无论它是什么...就像打开浏览器警报以保存文件时的浏览器一样。
我尝试使用response.headers["Content-type"]
来查找类型,但是在这种情况下,它始终是application/octet-stream
。
我该怎么办?
答案 0 :(得分:1)
您需要解析Content-Disposition
标头以获取正确的文件名。