Convert InputStream into FileInputStream

Question

I have read this post How to convert InputStream to FileInputStream on converting a InputStream into a FileInputStream. However, the answer does not work if you are using a resource that is in a jar file. Is there another way to do it.

I need to do this to get the FileChannel from a call to Object.class.getResourceAsStream(resourceName);

Why do you need it? In general, there may not be such an object. — Paŭlo Ebermann, Mar 18 '11 at 16:36
Note that addition to FileChannel.transferTo() -- which requires the source to be a FileChannel -- there is also FileChannel.transferFrom(), which requires the destination to be so. — Andy Thomas, Mar 18 '11 at 17:53

score 7 · Accepted Answer · answered Mar 18 '11 at 16:40

You can't, without basically writing to a file. Unless there's a file, there can't be a FileInputStream or a FileChannel. If at all possible, make sure your code is agnostic to the input source - design it in terms of InputStream and ByteChannel (or whatever kind of channel is most appropriate).

score 6 · Answer 2 · answered Mar 18 '11 at 16:41

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From the InputStream returned by Class.getResourceAsStream(), you can make a Channel with Channels.newChannel( InputStream ).

This is not the FileChannel you requested, but it is still a Channel. Is it sufficient to meet your needs?

answered Mar 18 '11 at 16:41

Andy Thomas

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Interesting, let me try this out – user489041 Mar 18 '11 at 17:01

score 0 · Answer 3 · answered Mar 18 '11 at 16:49

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If you really need a file and you know that the resource is not inside a jar or loaded remotely, then you can use getResource instead.

URL resourceLocation = Object.class.getResource(resourcePath);
if (resourceLocation == null) { throw new FileNotFoundException(resourcePath); }
File myFile = new File(resourceLocation.toURI());

If you don't absolutely need a FileChannel or can't make assumptions about how your classpath is laid out, then Andy Thomas-Cramer's solution is probably the best.

answered Mar 18 '11 at 16:49

Mike Samuel

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Thanks Mike, but the resource I need will be inside a JAR file. – user489041 Mar 18 '11 at 17:11
In that case, you can't do it without mounting the jar as a file system somehow. A file channel presupposes a file. cheers – Mike Samuel Mar 18 '11 at 18:59

Convert InputStream into FileInputStream

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