Develop Reference

How to make a copy of a file containing images and text using java

I have some word documents and excel sheets which has some images along with the file text content. I want to create a copy of that file and keep it at a specific location. I tried the following method which is creating file at specified location but the file is corrupted and cannot be read.
InputStream document = Thread.currentThread().getContextClassLoader().getResourceAsStream("upgradeworkbench/Resources/Upgrade_TD_Template.docx");
try {
OutputStream outStream = null;
Stage stage = new Stage();
stage.setTitle("Save");
byte[] buffer= new byte[document.available()];
document.read(buffer);
FileChooser fileChooser = new FileChooser();
fileChooser.setInitialFileName(initialFileName);
if (flag) {
fileChooser.getExtensionFilters().addAll(new FileChooser.ExtensionFilter("Microsoft Excel Worksheet", "*.xls"));
} else {
fileChooser.getExtensionFilters().addAll(new FileChooser.ExtensionFilter("Microsoft Word Document", "*.docx"));
}
fileChooser.setTitle("Save File");
File file = fileChooser.showSaveDialog(stage);
if (file != null) {
outStream = new FileOutputStream(file);
outStream.write(buffer);
// IOUtils.copy(document, outStream);
}
} catch (IOException ex) {
System.out.println(ex.getMessage());
}
Can anyone suggest me any different ways to get the proper file.
PS: I am reading the file using InputStream because it is inside the project jar.
PPS: I also tried Files.copy() but it didnt work.

I suggest you never trust on InputStream.available to know the real size of the input, because it just returns the number of bytes ready to be immediately read from the buffer. It might return a small number, but doesn't mean the file is small, but that the buffer is temporarily half-full.
The right algorithm to read an InputStream fully and write it over an OutputStream is this:
int n;
byte[] buffer=new byte[4096];
do
{
n=input.read(buffer);
if (n>0)
{
output.write(buffer, 0, n);
}
}
while (n>=0);

You can use the Files.copy() methods.
Copies all bytes from an input stream to a file. On return, the input stream will be at end of stream.

Use:
Files.copy(document, file.toPath(), StandardCopyOption.REPLACE_EXISTING);
As the class says, the second argument is a Path, not a File.
Generally, since this is 2015, use Path and drop File; if an API still uses File, make it so that it uses it at the last possible moment and use Path all the way.

java.io.FileNotFoundException: /u01/app/webapps/out/pj/Create.xlsx

I am trying to Compare last modified date of two excel files and replace the old file with new file.
In Scenario : When there is no file in the first place, so the code copies the file to that location and later reads it.
Issue is : It throws a FileNotFound exception when the excel file is not present on the server,even after writing the file to the
server(via code),but the file is not seen on the server. It works on
my machine(windows),but fails when deployed on server.
Again, it works like charm when the file is present on the server,while the old is being replaced by the new file.
Can you please help and explain on why its failing in the above scenario,and only on server ?
if(row.getValue("fileType").toString().equals("xlsx")&&checkindatefolder.after(localdate))
{
messagelist.add("we are going to get the replace file in the server");
InputStream inp=folder.getFile();
ZipInputStream izs = new ZipInputStream(inp);
ZipEntry e = null;
while ((e = izs.getNextEntry()) != null) {
System.out.println("e.isDirectory(): "+e.isDirectory());
if (!e.isDirectory()) {
filename=e.getName();
System.out.println("filename: "+filename);
FileOutputStream os=new FileOutputStream("path"+e.getName());
byte[] buffer = new byte[4096];
int read=0;
System.out.println("writing to file");
while ((read=izs.read(buffer))> 0) {
System.out.println("1111");
os.write(buffer,0,read);
}
System.out.println("writing to file complete");
inp.close();
os.flush();
os.close();
}
}

Do all parts of the path exist?
So in your example:
/u01/app/webapps/out/pj/Create.xlsx
Do all subdirectories exist?
/u01/app/webapps/out/pj
If not, than trying to write there might fail with a FileNotFoundException.
You should create the directories with Files.creatDirectories(Path) first.

How to download files using java and moving them to a directory?

I've looked around for some script or some tutorial for downloading file using Java, and then moving them to a different directory. I've seen a lot of questions similar to mine answered, but they all vary. Is there a definite way to do this? I currently learning Java and don't have enough experience make functioning script for this. Could anyone help?

I personally think this is the best way to download a file from the web.
When you download the file, it is stored in current running program, not on the hard drive, unless you specify it to.
URL url;
URLConnection con;
DataInputStream dis;
FileOutputStream fos;
byte[] fileData;
try {
url = new URL("http://website.com/file.pdf"); //File Location goes here
con = url.openConnection(); // open the url connection.
dis = new DataInputStream(con.getInputStream());
fileData = new byte[con.getContentLength()];
for (int q = 0; q < fileData.length; q++) {
fileData[q] = dis.readByte();
}
dis.close(); // close the data input stream
fos = new FileOutputStream(new File("/Users/kfang/Documents/Download/file.pdf")); //FILE Save Location goes here
fos.write(fileData); // write out the file we want to save.
fos.close(); // close the output stream writer
}
catch(Exception m) {
System.out.println(m);
}

zip the files which are present at one FTP location and copy to another FTP location directly

I want to create zip file of files which are present at one ftp location and Copy this zip file to other ftp location without saving locally.
I am able to handle this for small size of files.It works well for small size files 1 mb etc
But if file size is big like 100 MB, 200 MB , 300 MB then its giving error as,
java.io.FileNotFoundException: STOR myfile.zip : 550 The process cannot access the
file because it is being used by another process.
at sun.net.ftp.FtpClient.readReply(FtpClient.java:251)
at sun.net.ftp.FtpClient.issueCommand(FtpClient.java:208)
at sun.net.ftp.FtpClient.openDataConnection(FtpClient.java:398)
at sun.net.ftp.FtpClient.put(FtpClient.java:609)
My code is
URLConnection urlConnection=null;
ZipOutputStream zipOutputStream=null;
InputStream inputStream = null;
byte[] buf;
int ByteRead,ByteWritten=0;
***Destination where file will be zipped***
URL url = new URL("ftp://" + ftpuser+ ":" + ftppass + "#"+ ftppass + "/" +
fileNameToStore + ";type=i");
urlConnection=url.openConnection();
OutputStream outputStream = urlConnection.getOutputStream();
zipOutputStream = new ZipOutputStream(outputStream);
buf = new byte[size];
for (int i=0; i<li.size(); i++)
{
try
{
***Souce from where file will be read***
URL u= new URL((String)li.get(i)); // this li has values http://xyz.com/folder
/myPDF.pdf
URLConnection uCon = u.openConnection();
inputStream = uCon.getInputStream();
zipOutputStream.putNextEntry(new ZipEntry((String)li.get(i).substring((int)li.get(i).lastIndexOf("/")+1).trim()));
while ((ByteRead = inputStream .read(buf)) != -1)
{
zipOutputStream.write(buf, 0, ByteRead);
ByteWritten += ByteRead;
}
zipOutputStream.closeEntry();
}
catch(Exception e)
{
e.printStackTrace();
}
}
if (inputStream != null) {
try {
inputStream .close();
}
catch (Exception e) {
e.printStackTrace();
}
}
if (zipOutputStream != null) {
try {
zipOutputStream.close();
} catch (Exception e){
e.printStackTrace();
}
}
Can anybody let me know how I can avoid this error and handle large files

This is unrelated to file sizes; as the error says, you can't replace the file because some other process is currently locking it.
The reason why you see it more often with large files is because these take longer to transfer hence the chance of concurrent accesses is higher.
So the only solution is to make sure that no one uses the file when you try to transfer it. Good luck with that.
Possible other solutions:
Don't use Windows on the server.
Transfer the file under a temporary name and rename it when it's complete. That way, other processes won't see incomplete files. Always a good thing.
Use rsync instead of inventing the wheel again.

Back in the day, before we had network security, there were FTP servers that allowed 3rd party transfers. You could use site specific commands and send a file to another FTP server directly. Those days are long gone. Sigh.
Ok, maybe not long gone. Some FTP servers support the proxy command. There is a discussion here: http://www.math.iitb.ac.in/resources/manuals/Unix_Unleashed/Vol_1/ch27.htm

Java: Bzip2 library

I need to create Bzip2 archive.
A downloaded bzip2 library from 'Apache ant'.
I use class CBZip2OutputStream:
String s = .....
CBZip2OutputStream os = new CBZip2OutputStream(fos);
os.write(s.getBytes(Charset.forName("UTF-8")));
os.flush();
os.close();
(I didn't find any example how to use it, so I decided to use it in this way)
But it creates a corrupted archive on the disk.

You have to add BZip2 header (two bytes: 'B','Z') before writing the content:
//Write 'BZ' before compressing the stream
fos.write("BZ".getBytes());
//Write to compressed stream as usual
CBZip2OutputStream os = new CBZip2OutputStream(fos);
... the rest ...
Then, for instance, you can extract contents of your bzipped file with cat compressed.bz2 | bunzip2 > uncompressed.txt on a *nix system.

I have not found an example but in the end I understood how to use CBZip2OutputStream so here is one :
public void createBZipFile() throws IOException{
// file to zip
File file = new File("plane.jpg");
// fichier compresse
File fileZiped= new File("plane.bz2");
// Outputstream for fileZiped
FileOutputStream fileOutputStream = new FileOutputStream(fileZiped);
fileOutputStream.write("BZ".getBytes());
// we getting the data in a byte array
byte[] fileData = getArrayByteFromFile(file);
CBZip2OutputStream bzip = null;
try{
bzip = new CBZip2OutputStream(fileOutputStream );
bzip.write(fileData, 0, fileData.length);
bzip.flush() ;
bzip.close();
}catch (IOException ex) {
ex.printStackTrace();
}
fos.close();
}