如何使用Spring资源从远程URL读取zip文件？

Question

I'd like to use a zip file from remote URL inmemory only, using spring Resource class:

UrlResource url = new UrlResource("https://path.to.TheFile.zip");
System.out.println(url.getFile().lastModified()); //=0, so probably it did not work

Result:

Caused by: java.io.FileNotFoundException: https://path.to.TheFile.zip (The syntax for the filename, directory or media is invalid)
    at java.util.zip.ZipFile.open(Native Method) ~[?:1.7.0_51]
    at java.util.zip.ZipFile.<init>(ZipFile.java:215) ~[?:1.7.0_51]
    at java.util.zip.ZipFile.<init>(ZipFile.java:145) ~[?:1.7.0_51]
    at java.util.zip.ZipFile.<init>(ZipFile.java:159) ~[?:1.7.0_51]

What might be wrong here?

I cannot use url.getInputStream() as I cannot pass the IS into a ZipFile: ZipFile(url.getInputStream()) //error

Answer 1

According to getFile() 's javadoc : (emphasis mine)

This implementation returns a File reference for the underlying URL/URI, provided that it refers to a file in the file system .

You should use getInputStream() instead. You could then just store the byte array into a local in-memory data structure (for instance, using StreamUtils.copyToByteArray ) and process it (take a look at How can I convert byte array to ZIP file ).

In order to use ZipFile as suggested in the edited question, you'd need to write the byte array to a temporary local File (hint: File.createTempFile and FileCopyUtils.copy() ), and then instantiate ZipFile straight away from it.

Remember closing the Stream if necessary and deleting the temporary File when you're done with it.