我的parse_item回调中有以下代码:

sel = Selector(response)
item['name'] = sel.xpath('//div[@class="productDescriptionBlock"]/h2/text()').extract()[0]
return item

但我得到:
exceptions.UnicodeEncodeError: 'charmap' codec can't encode character u'\uff01' in position 271761: character maps to <undefined>

我也尝试添加UnicodeEncodeError,但仍然得到相同的错误。
Traceback (most recent call last):
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/twisted/internet/base.py", line 824, in runUntilCurrent
    call.func(*call.args, **call.kw)
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/twisted/internet/task.py", line 638, in _tick
    taskObj._oneWorkUnit()
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/twisted/internet/task.py", line 484, in _oneWorkUnit
    result = next(self._iterator)
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/scrapy/utils/defer.py", line 57, in <genexpr>
    work = (callable(elem, *args, **named) for elem in iterable)
--- <exception caught here> ---
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/scrapy/utils/defer.py", line 96, in iter_errback
    yield next(it)
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/scrapy/contrib/spidermiddleware/offsite.py", line 23, in process_spider_output
    for x in result:
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/scrapy/contrib/spidermiddleware/referer.py", line 22, in <genexpr>
    return (_set_referer(r) for r in result or ())
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/scrapy/contrib/spidermiddleware/urllength.py", line 33, in <genexpr>
    return (r for r in result or () if _filter(r))
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/scrapy/contrib/spidermiddleware/depth.py", line 50, in <genexpr>
    return (r for r in result or () if _filter(r))
  File "/home/scraper/.fakeroot/lib/python2.7/site-packages/scrapy/contrib/spiders/crawl.py", line 73, in _parse_response
    for request_or_item in self._requests_to_follow(response):
  File "/home/scraper/.fakeroot/lib/python2.7/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 "/home/scraper/.fakeroot/lib/python2.7/site-packages/scrapy/contrib/linkextractors/sgml.py", line 124, in extract_links
    ).encode(response.encoding)
  File "/home/scraper/.fakeroot/lib/python2.7/encodings/cp1252.py", line 12, in encode
    return codecs.charmap_encode(input,errors,encoding_table)
exceptions.UnicodeEncodeError: 'charmap' codec can't encode character u'\x99' in position 349751: character maps to <undefined>

最佳答案

我以前见过这个。如果我没有错,那么您正在使用规则的链接提取器中的restrict_xpaths参数。
可能的解决方案是:
避免对特定站点使用restrict_xpaths。这是因为页面内容包含声明的编码中未定义的字符。
识别无效字符并在规则作用于它们之前将其替换。不过,这可能很棘手。
使用此应答中的中间件将响应重新编码为其声明的编码:UnicodeEncodeError after setting restrict_xpaths settings

10-08 05:03