Question

我正在编写一些抓取代码并遇到如上所述的错误。我的代码正在关注。

# -*- coding: utf-8 -*-
import scrapy
from myproject.items import Headline


class NewsSpider(scrapy.Spider):
    name = 'IC'
    allowed_domains = ['kosoku.jp']
    start_urls = ['http://kosoku.jp/ic.php']

    def parse(self, response):
        """
        extract target urls and combine them with the main domain
        """
        for url in response.css('table a::attr("href")'):
            yield(scrapy.Request(response.urljoin(url), self.parse_topics))

    def parse_topics(self, response):
        """
        pick up necessary information
        """
        item=Headline()
        item["name"]=response.css("h2#page-name ::text").re(r'.*（インターチェンジ）')
        item["road"]=response.css("div.ic-basic-info-left div:last-of-type ::text").re(r'.*道$')
        yield item

当我在shell脚本上单独执行时，我可以得到正确的响应，但是一旦它进入程序并运行，它就不会发生。

    2017-11-27 18:26:17 [scrapy.core.scraper] ERROR: Spider error processing <GET http://kosoku.jp/ic.php> (referer: None)
Traceback (most recent call last):
  File "/Users/sonogi/envs/scrapy/lib/python3.5/site-packages/scrapy/utils/defer.py", line 102, in iter_errback
    yield next(it)
  File "/Users/sonogi/envs/scrapy/lib/python3.5/site-packages/scrapy/spidermiddlewares/offsite.py", line 29, in process_spider_output
    for x in result:
  File "/Users/sonogi/envs/scrapy/lib/python3.5/site-packages/scrapy/spidermiddlewares/referer.py", line 339, in <genexpr>
    return (_set_referer(r) for r in result or ())
  File "/Users/sonogi/envs/scrapy/lib/python3.5/site-packages/scrapy/spidermiddlewares/urllength.py", line 37, in <genexpr>
    return (r for r in result or () if _filter(r))
  File "/Users/sonogi/envs/scrapy/lib/python3.5/site-packages/scrapy/spidermiddlewares/depth.py", line 58, in <genexpr>
    return (r for r in result or () if _filter(r))
  File "/Users/sonogi/scraping/myproject/myproject/spiders/IC.py", line 16, in parse
    yield(scrapy.Request(response.urljoin(url), self.parse_topics))
  File "/Users/sonogi/envs/scrapy/lib/python3.5/site-packages/scrapy/http/response/text.py", line 82, in urljoin
    return urljoin(get_base_url(self), url)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/urllib/parse.py", line 424, in urljoin
    base, url, _coerce_result = _coerce_args(base, url)
  File "/opt/local/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/urllib/parse.py", line 120, in _coerce_args
    raise TypeError("Cannot mix str and non-str arguments")
TypeError: Cannot mix str and non-str arguments
2017-11-27 18:26:17 [scrapy.core.engine] INFO: Closing spider (finished)

我很困惑，并且非常感谢任何人的帮助！

Answer 1

根据Scrapy文档，您正在使用的.css(selector)方法返回SelectorList个实例。如果您想要url的实际（unicode）字符串版本，请调用extract()方法：

def parse(self, response):
    for url in response.css('table a::attr("href")').extract():
        yield(scrapy.Request(response.urljoin(url), self.parse_topics))

Answer 2

由于第15行的代码，您收到此错误。由于response.css('table a::attr("href")')返回类型为list的对象，因此您必须首先将url的类型从list转换为str，然后才能解析代码到另一个功能。此外，attr语法可能会导致您出错，因为正确的attr标签没有""，因此它不是a::attr("href")而是a::attr(href)。

因此，删除了上述两个问题后，代码将如下所示：

def parse(self, response):
        """
        extract target urls and combine them with the main domain
        """

        url = response.css('table a::attr(href)')
        url_str = ''.join(map(str, url))     #coverts list to str
        yield response.follow(url_str, self.parse_topics)

我如何修复＆＃34; TypeError：不能混合str和非str参数＆＃34;？

2 个答案: