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我是scrapy，python的初学者。我尝试在scrapinghub中部署spider代码，遇到如下错误。下面是代码。

import scrapy
from bs4 import BeautifulSoup,SoupStrainer
import urllib2
from scrapy.selector import Selector
from scrapy.http import HtmlResponse
import re
import pkgutil
from pkg_resources import resource_string
from tues1402.items import Tues1402Item

data = pkgutil.get_data("tues1402","resources/urllist.txt")
class SpiderTuesday (scrapy.Spider):     
    name = 'tuesday'
    self.start_urls = [url.strip() for url in data]
    def parse(self, response):
       story = Tues1402Item()
       story['url'] = response.url
       story['title'] = response.xpath("//title/text()").extract()
       return story

是我的 spider.py 代码

import scrapy
class Tues1402Item(scrapy.Item):
    # define the fields for your item here like:
    # name = scrapy.Field()
    title = scrapy.Field()
    url = scrapy.Field()

是 items.py 代码和

from setuptools import setup, find_packages
setup(
    name         = 'tues1402',
    version      = '1.0',
    packages     = find_packages(),
    entry_points = {'scrapy': ['settings = tues1402.settings']},
    package_data = {'tues1402':['resources/urllist.txt']},
    zip_safe = False,
)

是 setup.py 代码。

以下是错误。

Traceback（最近一次调用最后一次）：文件“/usr/local/lib/python2.7/site-packages/scrapy/core/engine.py”，第 126 行，在 _next_request request = next(slot.start_requests) 文件“/ usr/local/lib/python2.7/site-packages/scrapy/spiders/ init .py”，第 70 行，在 start_requests 中产生 self.make_requests_from_url(url) 文件“/usr/local/lib/python2.7/site- packages/scrapy/spiders/ init .py”，第 73 行，在 make_requests_from_url 返回 Request(url, dont_filter=True) 文件“ /usr/local/lib/python2.7/site-packages/scrapy/http/request/init。 py”，第 25 行，在init self._set_url(url) 文件“ /usr/local/lib/python2.7/site-packages/scrapy/http/request/init.py"，第 57 行，在 _set_url 中引发 ValueError('请求 url 中缺少方案：%s' % self._url) ValueError：请求 url 中缺少方案：h

先感谢您

4

1 回答 1

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您的错误意味着 urlh不是有效的 url。您应该打印出您的self.start_urls并查看那里有哪些网址，您很可能有一个字符串h作为您的第一个网址。

似乎您的蜘蛛在这里迭代文本而不是 URL 列表：

data = pkgutil.get_data("tues1402","resources/urllist.txt")
class SpiderTuesday (scrapy.Spider):     
    name = 'tuesday'
    self.start_urls = [url.strip() for url in data]

假设您在urllist.txt文件中存储带有一些分隔符的 url，您应该将其拆分：

# assuming file has url in every line
self.start_urls = [url.strip() for url in data.splitlines()]
于 2017-02-14T09:33:10.467 回答