为什么my_str.decode（“utf-8”）仍然失败？[重复]

Question

我相信unicode三明治。我用unicode三明治。那么，为什么我在字节字符串（py 2.7）上运行以下命令。。。

label = label.decode("utf-8")

我仍然得到一个错误：

Traceback (most recent call last):
  File "/usr/local/lib/python2.7/site-packages/celery/app/trace.py", line 385, in trace_task
    R = retval = fun(*args, **kwargs)
  File "/usr/local/lib/python2.7/site-packages/celery/app/trace.py", line 648, in __protected_call__
    return self.run(*args, **kwargs)
  File "/opt/celery/cl/scrapers/tasks.py", line 638, in update_docket_info_iquery
    d = update_docket_metadata(d, report.metadata)
  File "/usr/local/lib/python2.7/site-packages/juriscraper/pacer/case_query.py", line 166, in metadata
    self._get_label_value_pair(bold, True, field_names)
  File "/usr/local/lib/python2.7/site-packages/juriscraper/pacer/docket_report.py", line 233, in _get_label_value_pair

    label = label.decode("utf-8") <---- Shouldn't this work?

  File "/usr/local/lib/python2.7/encodings/utf_8.py", line 16, in decode
    return codecs.utf_8_decode(input, errors, True)
UnicodeEncodeError: 'ascii' codec can't encode character u'\xa0' in position 6: ordinal not in range(128)

而且，当我试图对崩溃的线路进行解码时，为什么会抛出一个unicodeincodeerror？

我很困惑。再一次

Answer 1

您的日志显示了答案：

UnicodeEncodeError: 'ascii' codec can't encode character u'\xa0' in position 6: ordinal not in range(128)

Python 2.7不能解码字符串中的字符，因为它是非ASCII字符。这里的解决方案是完全在Unicode中工作，或者先编码，然后用适当的编解码器解码。

问题可能重复：UnicodeEncodeError:“ascii”编解码器无法对位置20处的字符u'\xa0'进行编码：序号不在范围内（128）