Python UnicodeEncodeError：'ascii' 编解码器无法对位置 0 中的字符进行编码：序号不在范围内（128）
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Python UnicodeEncodeError: 'ascii' codec can't encode character in position 0: ordinal not in range(128)
提问by Bin
In Python 2.7, see the following error, when trying to cast type to ensure it matches the output schema.
在 Python 2.7 中，当尝试强制转换类型以确保它与输出模式匹配时，会看到以下错误。
UnicodeEncodeError: 'ascii' codec can't encode character in position 0: ordinal not in range(128) Tried to find why and reproduced the error in Jupiter. By simply typing in.
UnicodeEncodeError: 'ascii' codec can't encode character in position 0: ordinal not in range(128) 试图找出原因并在 Jupiter 中重现错误。只需输入即可。
str(u'\u2013')
What is the way to cast type to string that can handle this type of error? Thanks!
将类型转换为可以处理此类错误的字符串的方法是什么？谢谢！
回答by akhilsp
Try this:
尝试这个：
u'\u2013'.encode('utf-8')
回答by Bin
I will answer my own question. Found an duplicated question. stackoverflow.com/questions/9942594/
我会回答我自己的问题。发现一个重复的问题。stackoverflow.com/questions/9942594/
But for simplicity, here is an elegant solution that works well with my use case:
但为简单起见，这是一个适用于我的用例的优雅解决方案：
def safe_str(obj):
try: return str(obj)
except UnicodeEncodeError:
return obj.encode('ascii', 'ignore').decode('ascii')
return ""
safe_str(u'\u2013')
Or simply use:
或者简单地使用：
u'\u2013'.encode('ascii', 'ignore')
回答by Ashish Bainade
For version 2.7.x ,encoding is not set by default. Please use below code as a first line of program
对于 2.7.x 版本，默认情况下未设置编码。请使用以下代码作为程序的第一行
# -*- coding: utf-8 -*-
# Your code goes below this line
It should solve your problem.
它应该可以解决您的问题。
For python 3.x ,there is default encoding.hence there will be no issue of encoding.
对于python 3.x，有默认编码。因此不会有编码问题。
