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Reading a binary file from the file system as a BLOB to use in rhino with javascript

I'm planing to use SheetJS with rhino. And sheetjs takes a binary object(BLOB if i'm correct) as it's input. So i need to read a file from the system using stranded java I/O methods and store it into a blob before passing it to sheetjs. eg :-
var XLDataWorkBook = XLSX.read(blobInput, {type : "binary"});
So how can i create a BLOB(or appropriate type) from a binary file in java in order to pass it in.
i guess i cant pass streams because i guess XLSX needs a completely created object to process.

I found the answer to this by myself. i was able to get it done this way.
Read the file with InputStream and then write it to a ByteArrayOutputStream. like below.
ByteArrayOutputStream buffer = new ByteArrayOutputStream();
...
buffer.write(bytes, 0, len);
Then create a byte array from it.
byte[] byteArray = buffer.toByteArray();
Finally i did convert it to a Base64 String (which is also applicable in my case) using the "Base64.encodeBase64String()" method in apache.commons.codec.binary package. So i can pass Base64 String as a method parameter.
If you further need there are lot of libraries(3rd-party and default) available for Base64 to Blob conversion as well.

How to build FileInputStream object with byte array as a parameter

I have a zip file and after decoding it I get a byte array now I want to create a FileInputStream object with that byte[] object. I dont want to create a file instead pass data content do FileInputStream.
Is there any way ?
following is the code:
byte[] decodedHeaderFileZip = decodeHeaderZipFile(headerExportFile);
FileInputStream fileInputStream = new FileInputStream(decodedHeaderZipFileString);
EDIT:
I wanted to build a ZipInputStream object with a FileInputStream.

I have a zip file and after decoding it I get a byte array now I want to create a FileInputStream object with that byte[] object.
But you don't have a file. You have some data in memory. So a FileInputStream is inappropriate - there's no file for it to read from.
If possible, use a ByteArrayInputStream instead:
InputStream input = new ByteArrayInputStream(decodedHeaderFileZip);
Where possible, express your API in terms of InputStream, Reader etc rather than any specific implementation - that allows you to be flexible in which implementation you use. (What I mean is that where possible, make method parameters and return types InputStream rather than FileInputStream - so that callers don't need to provide the specific types.)
If you absolutely have to create a FileInputStream, you'll need to write the data to a file first.

How to convert byte array to file

I have connected to an ftp location using;
URL url = new URL("ftp://user:password#mydomain.com/" + file_name +";type=i");
I read the content into a byte array as shown below;
byte[] buffer = new byte[1024];
int count = 0;
while((count = fis.read(buffer)) > 0)
{
//check if bytes in buffer is a file
}
I want to be able to check if the bytes in buffer is a file without explicitly passing a specific file to write to it like;
File xfile= new File("dir1/");
FileOutputStream fos = new FileOutputStream(xfile);
fos.write(bytes);
if(xfile.isFile())
{
}
In an Ideal world something like this;
File xfile = new File(buffer);//Note: you cannot do this in java
if(xfile.isFile())
{
}
isFile() is to check if the bytes read from the ftp is file. I don't want to pass an explicit file name as I do not know the name of the file on the ftp location.
Any solutions available?

What is a file?
A computer file is a block of arbitrary information [...] which is available to a computer program and is usually based on some kind of durable storage. A file is durable in the sense that it remains available for programs to use after the current program has finished.
Your bytes that are stored in the byte array will be a part of a file if you write them on some kind of durable storage.
Sure, we often say that we read a file or write a file, but basically we read bytes from a file and write bytes to a file.
So we can't test a byte array whether it's content is a file or not. Simply because every byte array can be used to create a file (even an empty array).
BTW - the ftp server does not send a file, it (1) reads bytes and (2) a filename and (3) sends the bytes and (4) the filename so that a client can (5) read the bytes and (6) the filename and use both datasets to (7) create a file. The ftp server doesn't have to access a file, it can take bytes and names from a database or create both in memory...

I guess you cannot check if the byte[] array is a file or not. Why dont' you just use already written and tested library like maybe for example: http://commons.apache.org/net/

There is no way to do that easily.
A file is a byte array on a disk and a byte array will be a file if you write it to disk. There is no reliable way of telling what is in the data you just received, without parsing the data and checking if you can find a valid file header in it.

Where is isFile() file means the content fetched from from the ftp stream is a file.
The answer to that is simple. You can't do it because it doesn't make any sense.
What you have read from the stream IS a sequence of bytes stored in memory.
A file is a sequence of bytes stored on a disk (typically).
These are not the same thing. (Or if you want to get all theoretical / philosophical you have to answer the question "when is a sequence of bytes a file, and when is it not a file".
Now a more sensible question to ask might be:
How do I know if the stuff I fetched by FTP is the contents of a file on the FTP server.
(... as distinct from a rendering of a directory or something).
The answer is that you can't be sure if you fetched the file by opening an URLConnection to the FTP server ... like you have done. It is like asking "is '(123) 555-5555' a phone number?". It could be a phone number, or it could just be a sequence of characters that look like a phone number.

Java - directly reading a file without downloading and storing locally [duplicate]

This question already has answers here:
How do I read / convert an InputStream into a String in Java?
(62 answers)
Closed 8 years ago.
I want to directly read a file, put it into a string without storing the file locally. I used to do this with an old project, but I don't have the source code anymore. I used to be able to get the source of my website this way.
However, I don't remember if I did it by "InputStream to String array of lines to String", or if I directly read it into a String.
Was there a function for this, or am I remembering wrong?
(Note: this function would be the PHP equivalent of file_get_contents($path))

You need to use InputStreamReader to convert from a binary input stream to a Reader which is appropriate for reading text.
After that, you need to read to the end of the reader.
Personally I'd do all this with Guava, which has convenience methods for this sort of thing, e.g. CharStreams.toString(Readable).
When you create the InputStreamReader, make sure you supply the appropriate character encoding - if you don't, you'll get junk text out (just like trying to play an MP3 file as if it were a WAV, for example).

Check out apache-commons-io and for your use case FileUtils.readFileToString(File file)
(should not be to hard to get a File form the path).
You can use the library or have a look at the code - as this is open.

There is no direct way to read a File into a String.
But there is a quick alternative - read the File into a Byte array and convert it into a String.
Untested:
File f = new File("/foo/bar");
InputStream fStream = new FileInputStream(f);
ByteArrayOutputStream bStream = new ByteArrayOutputStream();
for(int data = fStream.read(); data > -1; data = fStream.read()) {
b.write(data);
}
String theResult = new String(bStream.toByteArray(), "UTF-8");

how to write a file without allocating the whole byte array into memory?

This is a newbie question, I know. Can you guys help?
I'm talking about big files, of course, above 100MB. I'm imagining some kind of loop, but I don't know what to use. Chunked stream?
One thins is for certain: I don't want something like this (pseudocode):
File file = new File(existing_file_path);
byte[] theWholeFile = new byte[file.length()]; //this allocates the whole thing into memory
File out = new File(new_file_path);
out.write(theWholeFile);
To be more specific, I have to re-write a applet that downloads a base64 encoded file and decodes it to the "normal" file. Because it's made with byte arrays, it holds twice the file size in memory: one base64 encoded and the other one decoded. My question is not about base64. It's about saving memory.
Can you point me in the right direction?
Thanks!

From the question, it appears that you are reading the base64 encoded contents of a file into an array, decoding it into another array before finally saving it.
This is a bit of an overhead when considering memory. Especially given the fact that Base64 encoding is in use. It can be made a bit more efficient by:
Reading the contents of the file using a FileInputStream, preferably decorated with a BufferedInputStream.
Decoding on the fly. Base64 encoded characters can be read in groups of 4 characters, to be decoded on the fly.
Writing the output to the file, using a FileOutputStream, again preferably decorated with a BufferedOutputStream. This write operation can also be done after every single decode operation.
The buffering of read and write operations is done to prevent frequent IO access. You could use a buffer size that is appropriate to your application's load; usually the buffer size is chosen to be some power of two, because such a number does not have an "impedance mismatch" with the physical disk buffer.

Perhaps a FileInputStream on the file, reading off fixed length chunks, doing your transformation and writing them to a FileOutputStream?

Perhaps a BufferedReader? Javadoc: http://download-llnw.oracle.com/javase/1.4.2/docs/api/java/io/BufferedReader.html

Use this base64 encoder/decoder, which will wrap your file input stream and handle the decoding on the fly:
InputStream input = new Base64.InputStream(new FileInputStream("in.txt"));
OutputStream output = new FileOutputStream("out.txt");
try {
byte[] buffer = new byte[1024];
int readOffset = 0;
while(input.available() > 0) {
int bytesRead = input.read(buffer, readOffset, buffer.length);
readOffset += bytesRead;
output.write(buffer, 0, bytesRead);
}
} finally {
input.close();
output.close();
}

You can use org.apache.commons.io.FileUtils. This util class provides other options too beside what you are looking for. For example:
FileUtils.copyFile(final File srcFile, final File destFile)
FileUtils.copyFile(final File input, final OutputStream output)
FileUtils.copyFileToDirectory(final File srcFile, final File destDir)
And so on.. Also you can follow this tut.