Python 设置限制路径设置后的UnicodeEncodeError

我不熟悉python和scrapy。将restrict\u xpath设置设置设置为“//table[@class=“lista”]”后，我收到了以下回溯。奇怪的是，通过使用其他xpath规则，爬虫程序可以正常工作

Traceback (most recent call last):
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/twisted/internet/base.py", line 800, in runUntilCurrent
    call.func(*call.args, **call.kw)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/twisted/internet/task.py", line 602, in _tick
    taskObj._oneWorkUnit()
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/twisted/internet/task.py", line 479, in _oneWorkUnit
    result = self._iterator.next()
  File "/Library/Python/2.7/site-packages/scrapy/utils/defer.py", line 57, in <genexpr>
    work = (callable(elem, *args, **named) for elem in iterable)
--- <exception caught here> ---
  File "/Library/Python/2.7/site-packages/scrapy/utils/defer.py", line 96, in iter_errback
    yield it.next()
  File "/Library/Python/2.7/site-packages/scrapy/contrib/spidermiddleware/offsite.py", line 23, in process_spider_output
    for x in result:
  File "/Library/Python/2.7/site-packages/scrapy/contrib/spidermiddleware/referer.py", line 22, in <genexpr>
    return (_set_referer(r) for r in result or ())
  File "/Library/Python/2.7/site-packages/scrapy/contrib/spidermiddleware/urllength.py", line 33, in <genexpr>
    return (r for r in result or () if _filter(r))
  File "/Library/Python/2.7/site-packages/scrapy/contrib/spidermiddleware/depth.py", line 50, in <genexpr>
    return (r for r in result or () if _filter(r))
  File "/Library/Python/2.7/site-packages/scrapy/contrib/spiders/crawl.py", line 73, in _parse_response
    for request_or_item in self._requests_to_follow(response):
  File "/Library/Python/2.7/site-packages/scrapy/contrib/spiders/crawl.py", line 52, in _requests_to_follow
    links = [l for l in rule.link_extractor.extract_links(response) if l not in seen]
  File "/Library/Python/2.7/site-packages/scrapy/contrib/linkextractors/sgml.py", line 124, in extract_links
    ).encode(response.encoding)
  File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/encodings/iso8859_2.py", line 12, in encode
    return codecs.charmap_encode(input,errors,encoding_table)
exceptions.UnicodeEncodeError: 'charmap' codec can't encode character u'\xbb' in position 686: character maps to <undefined>

---

对此错误的任何解释和帮助将不胜感激。谢谢。

这是由使用

的网页引起的错误实体，通过
lxml
转换为unicode字符
\xbb
，当您使用
restrict\u xpaths
参数时，链接提取器将内容编码为原始编码
iso8859-2
，该编码失败，因为
\xbb
在该编码中不是有效字符

此简单行再现了异常：

>>> u'\xbb'.encode('iso8859-2')
...
UnicodeEncodeError: 'charmap' codec can't encode character u'\xbb' in position 0: character maps to <undefined>

在您的设置中：

DOWNLOADER_MIDDLEWARES = {
    'myproject.middlewares.ForceUTF8Response': 100,
}

对不起，我发布了另一个问题的答案。@Rolando，这是否仍然适用于Scrapy v1？我不知道这是否应该应用于响应。text…@CloudRid你是对的。在Scrapy v1中，您应该使用
response.text
而不是
response.body\u作为unicode（）
（它是
response.text
的别名）。
# file: myproject/middlewares.py

class ForceUTF8Response(object):
    """A downloader middleware to force UTF-8 encoding for all responses."""
    encoding = 'utf-8'

    def process_response(self, request, response, spider):
        # Note: Use response.body_as_unicode() instead of response.text in in Scrapy <1.0.
        new_body = response.text.encode(self.encoding)
        return response.replace(body=new_body, encoding=self.encoding)

DOWNLOADER_MIDDLEWARES = {
    'myproject.middlewares.ForceUTF8Response': 100,
}