如何修复“类型错误:不能混合 str 和非 str 参数"? [英] How Can I Fix "TypeError: Cannot mix str and non-str arguments"?

问题描述

我正在编写一些抓取代码并遇到上述错误.我的代码如下.

I'm writing some scraping codes and experiencing an error as above. My code is following.

# -*- 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 脚本上单独执行它们时，我可以获得正确的响应，但是一旦它进入程序并运行，它就不会发生.

I can get the correct response when I do them individually on a shell script, but once it gets in a programme and run, it doesn't happen.

    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，感谢任何人的帮助！

I'm so confused and appreciate anyone's help upfront!

推荐答案

根据 Scrapy 文档，您使用的 .css(selector) 方法返回一个 SelectorList 实例.如果您想要 url 的实际(unicode)字符串版本，请调用 extract() 方法:

According to the Scrapy documentation, the .css(selector) method that you're using, returns a SelectorList instance. If you want the actual (unicode) string version of the url, call the extract() method:

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

这篇关于如何修复“类型错误:不能混合 str 和非 str 参数"?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！