Python Google App Engine：UnicodeDecodeError：“ascii”编解码器无法解码位置 48 中的字节 0xe2：序号不在范围内 (128)
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/20906948/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Google App Engine: UnicodeDecodeError: 'ascii' codec can't decode byte 0xe2 in position 48: ordinal not in range(128)
提问by Manas Chaturvedi
I'm working on a small application using Google App Engine which makes use of the Quora RSS feed. There is a form, and based on the input entered by the user, it will output a list of links related to the input. Now, the applications works fine for one letter queries and most of two-letter words if the words are separated by a '-'. However, for three-letter words and some two-letter words, I get the following error:
我正在使用 Google App Engine 开发一个使用 Quora RSS 提要的小型应用程序。有一个表单，根据用户输入的输入，它会输出与输入相关的链接列表。现在，如果单词由“-”分隔，应用程序可以很好地处理单字母查询和大多数双字母单词。但是，对于三个字母的单词和一些两个字母的单词，我收到以下错误：
UnicodeDecodeError: 'ascii' codec can't decode byte 0xe2 in position 48: ordinal not in range(128)
UnicodeDecodeError：“ascii”编解码器无法解码位置 48 中的字节 0xe2：序号不在范围内（128）
Here's my Python code:
这是我的 Python 代码：
import os
import webapp2
import jinja2
from google.appengine.ext import db
import urllib2
import re
template_dir = os.path.join(os.path.dirname(__file__), 'templates')
jinja_env = jinja2.Environment(loader = jinja2.FileSystemLoader(template_dir), autoescape=True)
class Handler(webapp2.RequestHandler):
def write(self, *a, **kw):
self.response.out.write(*a, **kw)
def render_str(self, template, **params):
t = jinja_env.get_template(template)
return t.render(params)
def render(self, template, **kw):
self.write(self.render_str(template, **kw))
class MainPage(Handler):
def get(self):
self.render("formrss.html")
def post(self):
x = self.request.get("rssquery")
url = "http://www.quora.com/" + x + "/rss"
content = urllib2.urlopen(url).read()
allTitles = re.compile('<title>(.*?)</title>')
allLinks = re.compile('<link>(.*?)</link>')
list = re.findall(allTitles,content)
linklist = re.findall(allLinks,content)
self.render("frontrss.html", list = list, linklist = linklist)
app = webapp2.WSGIApplication([('/', MainPage)], debug=True)
Here's the html code:
这是html代码：
<h1>Quora Live Feed</h1><br><br><br>
{% extends "rssbase.html" %}
{% block content %}
{% for e in range(1, 19) %}
{{ (list[e]) }} <br>
<a href="{{ linklist[e] }}">{{ linklist[e] }}</a>
<br><br>
{% endfor %}
{% endblock %}
采纳答案by Jon Wayne Parrott
Python is likelytrying to decode a unicode string into a normal str with the ascii codec and is failing. When you're working with unicode data you need to decode it:
Python可能会尝试使用 ascii 编解码器将 unicode 字符串解码为普通 str 并且失败。当您使用 unicode 数据时，您需要对其进行解码：
content = content.decode('utf-8')
回答by KimKha
In my AppEngine app, I convert it like:
在我的 AppEngine 应用程序中，我将其转换为：
content = unicode(content)
I think it more clear and easy to use.
我认为它更清晰，更易于使用。
