Python：生成器代码中的异常处理

Question

我有以下代码读取用户上传的文件并将其显示在预览页面上，预期的输入文件是纯文本但一些用户抱怨他们收到以下错误，

data = fileObj.read(4194304) 文件“/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/codecs.py”, 第 698 行，已读 返回 self.reader.read(size) 文件“/Library/Frameworks/Python.framework/Versions/3.6/lib/python3.6/codecs.py”， 第 501 行，已读 newchars, decodedbytes = self.decode(data, self.errors) UnicodeDecodeError: 'utf-8' codec can't decode byte 0xa9 in position 15: 无效的起始字节

读取文件的 Python 代码：

def gen():
    with codecs.open(file_path, 'r', encoding='utf-8') as fileObj:
        while True:
            # 4MB chunk (4 * 1024 * 1024 Bytes)
            data = fileObj.read(4194304)
            if not data:
                break
            yield data

return Response(gen(), mimetype='text/plain')

如何为生成器编写错误处理程序，它将编码切换为latin-1 并尝试再次读取文件而不是抛出异常？

更新容易失败的示例文件数据：

/*
    Copyright � 2017, J. Wayne Schneider - All Rights Reserved
    Unauthorized copying of this file, via any medium is strictly prohibited
    Proprietary and confidential
    Written by J. Wayne Schneider <jwaynes@gmail.com>, May 2017
*/

Answer 1

这将是我的看法。

def main():

    def gen(method):
        with open('path', encoding=method) as fileObj:
            while True:
                data = fileObj.read(4194304)
                if not data:
                    break
                yield data

    try:
        rr = [*gen('utf-8')]
    except UnicodeDecodeError:
        rr = [*(gen('latin-1')]
    return Response(rr, mimetype='text/plain')

---

def main():

    def gen():
        try:
            with open('path', encoding='utf-8') as fileObj:
                while True:
                    data = fileObj.read(4194304)
                    if not data:
                        break
                    yield data
        except (UnicodeDecodeError, UnicodeEncodeError, UnicodeError):
            with open('path', encoding='latin-1') as fileObj:
                while True:
                    data = fileObj.read(4194304)
                    if not data:
                        break
                    yield data

    return Response(gen('utf-8'), mimetype='text/plain')

这是带有try的版本；除了内置到 gen 中，不太可能抛出任何错误。

Answer 2

一个使用try的例子，除了处理错误：

def gen():
    with codecs.open(file_path, 'r', encoding='utf-8') as fileObj:
        while True:
            # 4MB chunk (4 * 1024 * 1024 Bytes)
            try:
                data = fileObj.read(4194304)
                if not data:
                    break
                yield data
            except UnicodeDecodeError:
                #do whatever you want on fail here
            except:
                raise ValueError('object could not be read, even after we caught the unicode error and did something else.')

    return Response(gen(), mimetype='text/plain')