python zlib如何解压多个对象

Question

我试图在我的文件上写很多 zlib 块，有没有办法从我的文件中获取所有未压缩的内容？从下面的示例代码中我只能得到我的第一个数据。预先感谢您的任何输入！

Python 3.6.8 |Anaconda, Inc.| (default, Dec 29 2018, 19:04:46) 
[GCC 4.2.1 Compatible Clang 4.0.1 (tags/RELEASE_401/final)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import zlib
>>> str1 = b'my first string'
>>> str2 = b'my second string'
>>> z1 = zlib.compress(str1)
>>> z2 = zlib.compress(str2)
>>> with open('test.z', 'wb') as fh:
...     fh.write(z1)
...     fh.write(z2)
... 
23
24
>>> p = open('test.z','rb').read()
>>> p
b'x\x9c\xcb\xadTH\xcb,*.Q(.)\xca\xccK\x07\x00.6\x05\xe6x\x9c\xcb\xadT(NM\xce\xcfKQ(.)\xca\xccK\x07\x003\xfb\x06:'
>>> zlib.decompress(p)
b'my first string'
>>> zlib.decompress(p)
b'my first string'

Answer 1

无法在单个函数或方法调用中解压缩串联的压缩流^*。但是，使用 zlib 包中的工具可以获得等效的功能。

>>> import zlib
>>> b1 = b'Here is some data'
>>> b2 = b'Here is some more data'
>>> stream = zlib.compress(b1) + zlib.compress(b2)


>>> while stream:
...     dco = zlib.decompressobj()
...     dec = dco.decompress(stream)
...     print(dec)
...     stream = dco.unused_data
... 
b'Here is some data'
b'Here is some more data'

zlib.decompressobj() creates an object which decompresses a single compressed stream - like zlib.decompress - and exposes the remaining bytes in the stream in its unused_data 属性。所以我们可以在一个循环中解压缩，直到整个连接的流都被解压缩。

^*至少我找不到

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