Question

我的代码中出现了这个常见错误：

Exception Type: UnicodeEncodeError
Exception Value:'ascii' codec can't encode character u'\xe9' in position 6: ordinal not in range(128)

为什么？因为我正在处理带有法语口音的名字。

这是我的代码：

if 'rechercheGED' in request.GET:

        query_social_number = request.GET.get('q1social')

        sort_params = {}

        Individu_Recherche.set_if_not_none(sort_params, 'NumeroIdentification__iexact', query_social_number)

        query_lastname_list = Individu_Recherche.Recherche_Get(Individu, sort_params)

        lastname = query_lastname_list.Nom
        firstname = query_lastname_list.Prenom
        NIU = query_lastname_list.NumeroIdentification

        title = str(lastname + "_" + firstname + "_" + NIU)

问题来自：firstname = query_lastname_list.Prenom

因为在我的情况下，名字是Jérôme

我尝试了一些事情：

1）在开头插入：

#-*- coding: utf-8 -*-
from __future__ import unicode_literals

2）使用firstname = query_lastname_list.Prenom.encode('utf-8')和firstname = query_lastname_list.Prenom.decode('utf-8')

但到目前为止，不可能删除此错误并使用重音处理数据。

你有什么想法吗？

编辑：

这是完整的追溯：

Environment:


Request Method: GET
Request URL: http://localhost:8000/Identification/Person/Research/?q1social=19910-00001-634239-2&rechercheGED=Rechercher

Django Version: 1.10.3
Python Version: 2.7.12
Installed Applications:
['Institution',
 'django.contrib.admin',
 'django.contrib.auth',
 'django.contrib.contenttypes',
 'django.contrib.sessions',
 'django.contrib.messages',
 'django.contrib.staticfiles',
 'bootstrapform',
 'django_countries',
 'debug_toolbar',
 'chartit',
 'Configurations',
 'Home',
 'Authentication',
 'Identity',
 'rest_framework']
Installed Middleware:
['django.contrib.sessions.middleware.SessionMiddleware',
 'django.middleware.locale.LocaleMiddleware',
 'django.middleware.common.CommonMiddleware',
 'django.middleware.csrf.CsrfViewMiddleware',
 'django.middleware.gzip.GZipMiddleware',
 'django.contrib.auth.middleware.AuthenticationMiddleware',
 'django.contrib.auth.middleware.SessionAuthenticationMiddleware',
 'django.contrib.messages.middleware.MessageMiddleware',
 'django.middleware.clickjacking.XFrameOptionsMiddleware',
 'django.middleware.security.SecurityMiddleware',
 'debug_toolbar.middleware.DebugToolbarMiddleware',
 'DatasystemsCORE.middleware.OnlineNowMiddleware']



Traceback:

File "/usr/local/lib/python2.7/site-packages/django/core/handlers/exception.py" in inner
  39.             response = get_response(request)

File "/usr/local/lib/python2.7/site-packages/django/core/handlers/base.py" in _legacy_get_response
  249.             response = self._get_response(request)

File "/usr/local/lib/python2.7/site-packages/django/core/handlers/base.py" in _get_response
  187.                 response = self.process_exception_by_middleware(e, request)

File "/usr/local/lib/python2.7/site-packages/django/core/handlers/base.py" in _get_response
  185.                 response = wrapped_callback(request, *callback_args, **callback_kwargs)

File "/usr/local/lib/python2.7/site-packages/django/contrib/auth/decorators.py" in _wrapped_view
  23.                 return view_func(request, *args, **kwargs)

File "/Users/valentinjungbluth/Desktop/Django/DatasystemsCORE/DatasystemsCore/DatasystemsCORE/Identity/views.py" in IdentityIndividuResearching
  454.         title = str(lastname + "_" + firstname + "_" + NIU)

Exception Type: UnicodeEncodeError at /Identification/Person/Research/
Exception Value: 'ascii' codec can't encode character u'\xe9' in position 6: ordinal not in range(128)

Answer 1

由于您使用的是Python 2.x，问题确实在于回溯的最后一行。

title = unicode(str(lastname + "_" + firstname + "_" + NIU))

lastname，firstname或NIU包含无法用7位ASCII表示的字符，这是str所做的（在Python 2中）。< / p>

Django为这些类型的字符串强制转换提供了有用的函数force_text和force_bytes，并且在执行此类操作时使用字符串插值而不是+也是个好主意。：

from django.utils.text import force_text

title = force_text('%s_%s_%s' % (lastname, firstname, NIU))

Answer 2

在Python 3中，您的title = str(lastname + "_" + firstname + "_" + NIU)会以您期望的方式运行。在Python 2中，您正在混合Unicode和ASCII，因此Python“有帮助”尝试将Unicode字符串转换为ASCII，以便将它们一起添加，当然如果Unicode无法用ASCII表示，则会失败。

解决此问题的一种简单方法是将所有内容都设为Unicode。例如，

firstname = u'Jérôme'
lastname = u'Pécresse'
title = lastname + u'_' + firstname
print title

<强>输出

Pécresse_Jérôme

我强烈建议您迁移到Python 3，一旦您习惯了不同（但更优越）的文本和字节处理，它将使事情变得更加愉快。

与此同时，您可能会发现这篇文章很有用：Pragmatic Unicode，由SO资深人士Ned Batchelder撰写。

Django问题：UnicodeEncodeError

2 个答案: