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