Java：从字节数组获取ZipFile的任何方法（或带有直接getEntry方法的任何方法）？

Question

I have the contents of a zip file in a byte array. The file contains a number of entries (typically about 12), but I only care about three of them.

I would like to somehow get this into a ZipFile object, so I can pull those specific three ZipEntry s out using ZipFile.getEntry . I'm open to using something other than ZipFile that has a similar look-up-by-name method like getEntry .

My initial investigation suggests that I'm out of luck. ZipFile requires a real file in the file subsystem (which I cannot and do not want to access) and so I can't get there from here, and no means other than ZipFile exists that allows extracting particular entries by name; but I wanted to check. In languages like C# and Python, this is pretty straightforward (in C# I go from byte array to MemoryStream to ZipArchive ; in Python I just wrap it in StringIO and treat like a file), so I wanted to make sure I'm not missing something obvious.

My Plan B is to use ZipInputStream and repeated calls to getNextEntry to go through all dozen or so entries, and throw away all except the three I care about, but that just smells bad to me.

Answer 1

A ZipInputStream can be instantiated for any InputStream ... including a ByteArrayInputStream .

Apart from that you are out of luck ... if you stick with Java SE classes.

The root of the problem (from an API design perspective) is that ZipFile is a wrapper for functionality that is implemented in native code. The native code opens the input stream for itself, and it uses a native filename / pathname.

The main reason for a native ZIP implementation that works that way is that the JVM needs to load code from ZIP files as part of the bootstrap procedures. This happens before the native implementation has loaded classes such as InputStream . Indeed, it has to.

There are a number of 3rd party libraries. Start by reading this Q&A - What is a good Java library to zip/unzip files?