Develop Reference

Facing issue while chunking then merging the jar files

I have one jar file for example apache-cassandra-3.11.6.jar.
Firstly i split/chunked into mutiple jars like below :
apache-cassandra1.jar
apache-cassandra2.jar
apache-cassandra3.jar
apache-cassandra4.jar
apache-cassandra5.jar
apache-cassandra6.jar
Then i reassemble them again into new Jar file i.e apache-cassandra_Merged.jar.
Now the problem comes.
When i compare the original jar file i.e apache-cassandra-3.11.6.jar with new Jar file i.e apache-cassandra_Merged.jar. then it is not matching.
The newly created jar file which is apache-cassandra_Merged.jar, it's size also reduced.
Please find below my code for your reference :
/// Chunking/spliting into mutiple jars
Path path = Paths.get("/Original_Jar/apache-cassandra-3.11.6.jar");
byte [] data = Files.readAllBytes(path); // Will read all bytes at once
Now divide total bytes into equal part and then write in each small jars one by one.
int count = 0;
for(byte[] rangeData : Arrays.copyOfRange(data, rangeSTART, rangeEND)){
FileOutputStream fileOutputStream1 = new FileOutputStream("/Cassandra_Image/Chunked_Jar/apache-cassandra"+count+".jar");
fileOutputStream1.write(rangeData);
}
//Merging back to one jar
For merging i used the same way. Created array of byte for each small/chunked jars and written into FileOutputStream("/Merged_Jar/apache-cassandra_Merged.jar") one by one.
Please let me know if i should use some other method/algorithm to split jar and reassemble it again which will make sure the originality of data after chunking and merging as well.
Note : Actually i want to transfer the jars to any server/directory where i should transfer a jar with limited size so for big size jars i need to split into small jars and send them one by one and then again reassemble them in target directory/place and it should be as original jar.
Thanks in advance.

This may not be the answer, but I provide as an information for you. Java also provides pack format where you can compress the jar files and then you can uncompress using unpack.
The tool is called pack200.
How to compress
<java_location>...\jre\lib>pack200 -J-Xmx256m small.jar.gz big.jar
How to uncompress
<java_location>...\jre\lib>unpack200 small.jar.gz big.jar
You can refer the following links.
https://docs.oracle.com/javase/1.5.0/docs/tooldocs/share/pack200.html
https://docs.oracle.com/javase/7/docs/technotes/tools/share/unpack200.html

I am able to solve the issue with shell scripting.
Written below code in my shell script file and run through my java code.
split -b 1000000 src.jar target.jar
cat src.jaraa src.jarab src.jarac src.jarad src.jarae > merged.jar
And compare with any algorithm like sha256 checksum will work fine and it shows equal. and size also equal.

create zip file without writing to disk

I am working on a Springboot application that has to return a zip file to a frontend when the user downloads some report. I want to create a zip file without writing the zip file or the original files to disk.
The directory I want to zip contains other directories, that contain the actual files. For example, dir1 has subDir1 and subDir2 inside, subDir1 will have two file subDir1File1.pdf and subDir1File2.pdf. subDir2 will also have files inside.
I can do this easily by creating the physical files on the disk. However, I feel it will be more elegant to return these files without writing to disk.

You would use ByteArrayOutputStream if the scope was to write to memory. In essence, the zip file would be entirely contained in memory, so be sure that you don't risk to have too many requests at once and that the file size is reasonable in size! Otherwise this approach can seriously backfire!

You can use following snippet :
public static byte[] zip(final String str) throws IOException {
if (StringUtils.isEmpty(str)) {
throw new IllegalArgumentException("Cannot zip null or empty string");
}
ByteArrayOutputStream bos = new ByteArrayOutputStream();
try (GZIPOutputStream gos = new GZIPOutputStream(bos)) {
gos.write(str.getBytes(StandardCharsets.UTF_8));
}
return bos.toByteArray();
}
But as stated in another answer, make sure you are not risking your program too much by loading everything into your java memory.

Please note that you should stream whenever possible. In your case, you could write your data to https://docs.oracle.com/javase/8/docs/api/index.html?java/util/zip/ZipOutputStream.html.
The only downside of this appproach is: the client won't be able to show a download status bar, because the server will not be able to send the "Content-length" header. That's because the size of a ZIP file can only be known after it has been generated, but the server needs to send the headers first. So - no temporary zip file - no file size beforehand.
You are also talking about subdirectories. This is just a naming issue when dealing with a ZIP stream. Each zip item needs to be named like this: "directory/directory2/file.txt". This will produce subdirectories when unzipping.

How to change a value of a file using Java IO function

Can anyone help me how to change a line in a file using java IO function. The file is located at SDCard. I have tried some solution from different blog, but failed.
I need to change one attribute wpa_oper_channel=1 to 2,3,4..... as user demand in the file sdcard/sample.txt.
I have also tried using SED command, but still not working. Please suggest if there any solution using java IO function.
The Command I have used using SED :
sed -i 's/wpa_oper_channel=[0-9]\\+/wpa_oper_channel=7/' sample.txt

If your configuration file is in the form of a Properties file. This means each line is in a format of key=value. You can easily read, modify and write it. Here is an example, I assume you have no problem with reading and writing a file through streams.
File sdcard = Environment.getExternalStorageDirectory();
File file = new File(sdcard,"sample.txt");
InputStream is = new FileInputStream(file); // Read your existing file
Properties props = new Properties();
props.load(is);
props.setProperty("wpa_oper_channel", "4");
OutputStream out = new FileOutputStream(file); // Your output file stream
prop.store(output, null); // Save your properties file without header
By doing this, you may lose if there is another type of line like comments or else.
Update
I updated to source code. But still you need to get read and write permission.
<uses-permission android:name="android.permission.WRITE_EXTERNAL_STORAGE" />
I haven't test the code, It will work with some exception handling. If it is not work please specify your error.

You just have to
read the file's content into memory (there are millions of examples on how to do that with java),
change the content (finding the position could be done using e.g. indexOf("wpa_oper_channel="))
write the content back to the file (there are millions of examples for that too)

What android do you use?
From android 4.4+ user haven't access to edit and remove files from SD Card. You must copy them to external storage and use there. See this link - http://code.google.com/p/android/issues/detail?id=67570

Java File - Append pre-defined filename to a user-defined directory

I have a program that creates multiple output files e.g. daily_results.txt, overall_results.txt etc.
I want to allow the user to specify the directory that these files will be saved to using JFileChooser.
So if the user selected the directory they wanted their output to be saved to as "C:\temp\". What is the best way to append daily_results.txt to that file object. Is there a more elegant way to do this other than:
File file = new File(userDirectory.getPath() + "daily_results.txt");
Any ideas?

Apologies!
I think this can quite easily be accomplished with the JFileChoosers setSelectedFile method.

How to extract a single file from a remote archive file?

Given
URL of an archive (e.g. a zip file)
Full name (including path) of a file inside that archive
I'm looking for a way (preferably in Java) to create a local copy of that file, without downloading the entire archive first.
From my (limited) understanding it should be possible, though I have no idea how to do that. I've been using TrueZip, since it seems to support a large variety of archive types, but I have doubts about its ability to work in such a way. Does anyone have any experience with that sort of thing?
EDIT: being able to also do that with tarballs and zipped tarballs is also important for me.

Well, at a minimum, you have to download the portion of the archive up to and including the compressed data of the file you want to extract. That suggests the following solution: open a URLConnection to the archive, get its input stream, wrap it in a ZipInputStream, and repeatedly call getNextEntry() and closeEntry() to iterate through all the entries in the file until you reach the one you want. Then you can read its data using ZipInputStream.read(...).
The Java code would look something like this:
URL url = new URL("http://example.com/path/to/archive");
ZipInputStream zin = new ZipInputStream(url.getInputStream());
ZipEntry ze = zin.getNextEntry();
while (!ze.getName().equals(pathToFile)) {
zin.closeEntry(); // not sure whether this is necessary
ze = zin.getNextEntry();
}
byte[] bytes = new byte[ze.getSize()];
zin.read(bytes);
This is, of course, untested.

Contrary to the other answers here, I'd like to point out that ZIP entries are compressed individually, so (in theory) you don't need to download anything more than the directory and the entry itself. The server would need to support the Range HTTP header for this to work.
The standard Java API only supports reading ZIP files from local files and input streams. As far as I know there's no provision for reading from random access remote files.
Since you're using TrueZip, I recommend implementing de.schlichtherle.io.rof.ReadOnlyFile using Apache HTTP Client and creating a de.schlichtherle.util.zip.ZipFile with that.
This won't provide any advantage for compressed TAR archives since the entire archive is compressed together (beyond just using an InputStream and killing it when you have your entry).

Since TrueZIP 7.2, there is a new client API in the module TrueZIP Path. This is an implementation of an NIO.2 FileSystemProvider for JSE 7. Using this API, you can access HTTP URI as follows:
Path path = new TPath(new URI("http://acme.com/download/everything.tar.gz/README.TXT"));
try (InputStream in = Files.newInputStream(path)) {
// Read archive entry contents here.
...
}

I'm not sure if there's a way to pull out a single file from a ZIP without downloading the whole thing first. But, if you're the one hosting the ZIP file, you could create a Java servlet which reads the ZIP file and returns the requested file in the response:
public class GetFileFromZIPServlet extends HttpServlet{
#Override
public void doGet(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException{
String pathToFile = request.getParameter("pathToFile");
byte fileBytes[];
//get the bytes of the file from the ZIP
//set the appropriate content type, maybe based on the file extension
response.setContentType("...");
//write file to the response
response.getOutputStream().write(fileBytes);
}
}