Ansi到UTF-8使用python导致错误 [英] Ansi to UTF-8 using python causing error

问题描述

在我尝试编写一个将Ansi转换为UTF-8的python程序时，我发现这个

如何使用Python将UTF-8转换为ANSI

它将UTF-8转换为Ansi。

我以为只要扭转订单就可以了。所以我编码了

  file_path_ansi =input.txt
 file_path_utf8 =output.txt
 b $ b #open并编码原始内容
 file_source = open（file_path_ansi，mode ='r'，encoding ='latin-1'，errors ='ignore'）
 file_content = file_source.read ）
 file_source.close 
 
 #write 
 file_target = open（file_path_utf8，mode ='w'，encoding ='utf-8'）
 file_target.write file_content）
 file_target.close 
  

但它会导致错误。

  TypeError：file<>最多取3个参数< 4给定> 
  

所以我改变了

  file_source = open（file_path_ansi，mode ='r'，encoding ='latin-1'，errors ='ignore'）
   
 
到
  file_source = open（file_path_ansi，mode ='r'，encoding = 'latin-1'）
  
然后会导致另一个错误：
  TypeError：'encoding'是此函数的无效关键字argumentemtn 
  
我应该如何修复我的代码来解决这个问题？
解决方案
 3版本的  open（）函数，在Python 2.在主要版本之间，I / O支持被大修，支持更好的编码和解码。
 
 
 
 Python 2中的版本为  io.open（） 。
 
 
 
我会使用  shutil.copyfileobj（）函数进行复制，因此您不必将整个文件读入内存：
 $ igb 
 $ b 
  import io 
 
使用io.open（file_path_ansi，encoding = 'latin-1'，errors ='ignore'）as source：
 with io.open（file_path_utf8，mode ='w'，encoding ='utf-8'）as target：
 shutil.copyfileobj （源，目标）
  
大多数谈到ANSI的人都指的是 Windows代码页的一个;你可能真的有一个文件在CP（代码页）1252，这是几乎，但不完全相同的事情 ISO-8859-1（拉丁语1）。如果是，使用 cp1252 而不是 latin-1 作为编码参数。
 
While I was trying to write a python program that converts Ansi to UTF-8, I found this

How can I convert UTF-8 to ANSI in Python

which converts UTF-8 to Ansi. 

I thought it will just work by reversing the order. So I coded
file_path_ansi = "input.txt"
file_path_utf8 = "output.txt"

#open and encode the original content
file_source = open(file_path_ansi, mode='r', encoding='latin-1', errors='ignore')
file_content = file_source.read()
file_source.close

#write 
file_target = open(file_path_utf8, mode='w', encoding='utf-8')
file_target.write(file_content)
file_target.close
But it causes error.
TypeError: file<> takes at most 3 arguments <4 given>
So I changed 
file_source = open(file_path_ansi, mode='r', encoding='latin-1', errors='ignore')
to 
file_source = open(file_path_ansi, mode='r', encoding='latin-1')
Then it causes another error:
TypeError: 'encoding' is an invalid keyword arguemtn for this function
How should I fix my code to solve this problem?
解决方案
You are trying to use the Python 3 version of the open() function, on Python 2. Between the major versions, I/O support was overhauled, supporting better encoding and decoding.

You can get the same new version in Python 2 as io.open() instead.

I'd use the shutil.copyfileobj() function to do the copying, so you don't have to read the whole file into memory:
import io

with io.open(file_path_ansi, encoding='latin-1', errors='ignore') as source:
    with io.open(file_path_utf8, mode='w', encoding='utf-8') as target:
        shutil.copyfileobj(source, target)
Be careful though; most people talking about ANSI refer to one of the Windows codepages; you may really have a file in CP (codepage) 1252, which is almost, but not quite the same thing as ISO-8859-1 (Latin 1). If so, use cp1252 instead of latin-1 as the encoding parameter.

                        
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