如何将 ZipFile 对象转换为支持缓冲区 API 的对象？

Question

我有一个 ZipFile 对象，我需要将其转换为可与缓冲区一起使用的对象 api。上下文是我正在尝试使用一个 API，它表示它需要一个类型为 string($binary) 的文件。我该怎么做呢？我知道这是完全错误的，但这是我的代码：

    def create_extension_zip_file(self, path_to_extension_directory, directory_name):
    zipObj = ZipFile("static_extension.zip", "w")
    with zipObj:
        # Iterate over all the files in directory
        for folderName, subfolders, filenames in os.walk(path_to_extension_directory):
            for filename in filenames:
                # create complete filepath of file in directory
                filePath = os.path.join(folderName, filename)
                with open(filename, 'rb') as file_data:
                    bytes_content = file_data.read()
                # Add file to zip
                zipObj.write(bytes_content, basename(filePath))
    return zipObj

Answer 1

或者如果 API 需要一个 file-like 对象，您可以在创建压缩文件时传递一个 BytesIO 实例并将其传递给 API

import io

def create_extension_zip_file(self, path_to_extension_directory, directory_name):
    buf = io.BytesIO()
    zipObj = ZipFile(buf, "w")
    with zipObj:
        # Iterate over all the files in directory
        for folderName, subfolders, filenames in os.walk(path_to_extension_directory):
            for filename in filenames:
                # create complete filepath of file in directory
                filePath = os.path.join(folderName, filename)
                with open(filename, 'rb') as file_data:
                    bytes_content = file_data.read()
                # Add file to zip
                zipObj.write(bytes_content, basename(filePath))
    # Rewind the buffer's file pointer (may not be necessary)
    buf.seek(0)
    return buf

如果 API 需要一个 bytes 实例，您可以在写入后以二进制模式打开 zip 文件，然后传递 bytes .

with open('static_extension.zip', 'rb') as f:
    bytes_ = f.read()

如何将 ZipFile 对象转换为支持缓冲区 API 的对象？

How do I convert a ZipFile object to an object supporting the buffer API?

python

object

binary

python-zipfile