Question

我想使用scrapy抓取我在Downloads文件夹中找到的本地xml文件，使用xpath提取相关信息。

将scrapy简介用作guide

2016-01-24 12:38:53 [scrapy] DEBUG: Retrying <GET file://home/sayth/Downloads/20160123RAND0.xml> (failed 2 times): [Errno 2] No such file or directory: '/sayth/Downloads/20160123RAND0.xml'
2016-01-24 12:38:53 [scrapy] DEBUG: Gave up retrying <GET file://home/sayth/Downloads/20160123RAND0.xml> (failed 3 times): [Errno 2] No such file or directory: '/sayth/Downloads/20160123RAND0.xml'
2016-01-24 12:38:53 [scrapy] ERROR: Error downloading <GET file://home/sayth/Downloads/20160123RAND0.xml>

我已经尝试了下面的几个版本但是我无法获得启动网址来接受我的文件。

# -*- coding: utf-8 -*-
import scrapy


class MyxmlSpider(scrapy.Spider):
    name = "myxml"
    allowed_domains = ["file://home/sayth/Downloads"]
    start_urls = (
        'http://www.file://home/sayth/Downloads/20160123RAND0.xml',
    )

    def parse(self, response):
        for file in response.xpath('//meeting'):
            full_url = response.urljoin(href.extract())
            yield scrapy.Request(full_url, callback=self.parse_question)

    def parse_xml(self, response):
        yield {
            'name': response.xpath('//meeting/race').extract()
        }

只是为了确认我确实在那个位置有文件

sayth@sayth-HP-EliteBook-2560p : ~/Downloads
[0] % ls -a
.                                                              Building a Responsive Website with Bootstrap [Video].zip
..                                                             codemirror.zip
1.1 Situation Of Long Term Gain.xls                            Complete-Python-Bootcamp-master.zip
2008 Racedata.xls                                              Cox Plate 2005.xls
20160123RAND0.xml

Answer 1

根本不要指定allowed_domains并在协议后使用 3斜杠：

start_urls = ["file:///home/sayth/Downloads/20160123RAND0.xml"]

Answer 2

必须使用文件的绝对路径通过file://协议指定本地文件。
我个人建议为此使用pathlib，而不要自己用字符串指定绝对值。

这是一个用法示例

import pathlib

start_urls = [
  pathlib.Path(os.path.abspath('20160123RAND0.xml')).as_uri()
]

as_uri()方法将路径转换为file:// uri

使用Scrapy爬网本地XML文件 - 启动URL本地文件地址

2 个答案: