Develop Reference

java unknown protocol: e when downloading a file

I'm a beginner to java file handling. I tired to get a bin file (en-parser-chunking.bin) from my hard disk partition to my web application. So far I have tried below code and it gives me the output in my console below.
unknown protocol: e
these are the code samples I have tried so far
//download file
public void download(String url, File destination) throws IOException {
URL website = new URL(url);
ReadableByteChannel rbc = Channels.newChannel(website.openStream());
FileOutputStream fos = new FileOutputStream(destination);
fos.getChannel().transferFrom(rbc, 0, Long.MAX_VALUE);
}
public void parserAction() throws Exception {
//InputStream is = new FileInputStream("en-parser-chunking.bin");
File modelFile = new File("en-parser-chunking.bin");
if (!modelFile.exists()) {
System.out.println("Downloading model.");
download("E:\\Final Project\\Softwares and tools\\en-parser-chunking.bin", modelFile);
}
ParserModel model = new ParserModel(modelFile);
Parser parser = ParserFactory.create(model);
Parse topParses[] = ParserTool.parseLine(line, parser, 1);
for (Parse p : topParses){
//p.show();
getNounPhrases(p);
}
}
getting a file in this way is possible or I have done it wrong ?
note - I need to get this from my hard disk. not download from the internet

the correct URL for a local file is:
file://E:/Final Project/Softwares and tools/en-parser-chunking.bin
where file is the protocol.
You can also you:
new File("E:/Final Project/Softwares and tools/en-parser-chunking.bin").toURL()
to create a URL from your file.
I also recomment to use slash as file seperator instead of backslash

Getting strange structure file when zipping a directory using Java

I wanted to zip a directory with files and subdirectories in it. I did this and worked fine but I am getting and unusual and curious file structure (At least I see it that way).
This is the created file: When I click on it, I see an "empty" directory like this: but when I unzip this I see this file structure (Not all the names are exacly as they are showed in the image below):
|mantenimiento
|Carpeta_A
|File1.txt
|File2.txt
|Carpeta_B
|Sub_carpetaB
|SubfileB.txt
|Subfile1B.txt
|Subfile2B.txt
|File12.txt
My problem somehow is that the folder "mantenimiento" is where I am zippping from (the directory which I want to zip) and I dont want it to be there, so when I unzip the just created .zip file I want it with this file structure (which are the files and directories inside "mantenimiento" directory): and the other thing is when I click on the .zip file I want to see the files and directories just like the image showed above.
I dont know what's wrong with my code, I have searched but haven't found a reference to what my problem might be.
Here's my code:
private void zipFiles( List<File> files, String directory) throws IOException
{
ZipOutputStream zos = null;
ZipEntry zipEntry = null;
FileInputStream fin = null;
FileOutputStream fos = null;
BufferedInputStream in = null;
String zipFileName = getZipFileName();
try
{
fos = new FileOutputStream( File.separatorChar + zipFileName + EXTENSION );
zos = new ZipOutputStream(fos);
byte[] buf = new byte[1024];
int len;
for(File file : files)
{
zipEntry = new ZipEntry(file.toString());
fin = new FileInputStream(file);
in = new BufferedInputStream(fin);
zos.putNextEntry(zipEntry);
while ((len = in.read(buf)) >= 0)
{
zos.write(buf, 0, len);
}
}
}
catch(Exception e)
{
System.err.println("No fue posible zipear los archivos");
e.printStackTrace();
}
finally
{
in.close();
zos.closeEntry();
zos.close();
}
}
Hope you guys can give me a hint about what I am doing wrong or what I am missing.
Thanks a lot.
Btw, the directory i am giving to the method is never used. The other parameter i am giving is a list of files which contains all the files and directories from the C:\mantenimiento directory.

I once had a problem with windows and zip files, where the created zip did not contain the entries for the folders (i.e. /, /Carpeta_A etc) only the file entries. Try adding ZipEntries for the folders without streaming content.
But as alternative to the somewhat bulky Zip API of Java you could use Filesystem (since Java7) instead. The following example is for Java8 (lambda):
//Path pathToZip = Paths.get("path/to/your/folder");
//Path zipFile = Paths.get("file.zip");
public Path zipPath(Path pathToZip, Path zipFile) {
Map<String, String> env = new HashMap<String, String>() {{
put("create", "true");
}};
try (FileSystem zipFs = FileSystems.newFileSystem(URI.create("jar:" + zipFile.toUri()), env)) {
Path root = zipFs.getPath("/");
Files.walk(pathToZip).forEach(path -> zip(root, path));
}
}
private static void zip(final Path zipRoot, final Path currentPath) {
Path entryPath = zipRoot.resolve(currentPath.toString());
try {
Files.createDirectories(entryPath.getParent());
Files.copy(currentPath, entryPath);
} catch (IOException e) {
throw new RuntimeException(e);
}
}

Create file in jsp servlet

When I create a file in java servlet, I can't find that file for opening. This is my code in servlet:
FileOutputStream fout;
try {
fout = new FileOutputStream("title.txt");
new PrintStream(fout).println(request.getParameter("txttitle"));
fout.close();
System.out.println(request.getParameter("txttitle"));
} catch (Exception e) {
System.out.println("I can't create file!");
}
Where I can find that file?

if you create file first as in
File f = new File("title.txt");
fout = new FileOutputStream(f);
then you use getAbsolutePath to return the location of where it has been created
System.out.println (f.getAbsolutePath());

Since you have'nt specified any directory for the file, it will be placed in the default directory of the process that runs your servlet container.
I would recommand you to always specify the full path of your your file when doing this kind of things.
If you're running tomcat, you can use System.getProperty("catalina.base") to get the path of the tomcat base directory. This can sometimes help.

Create a file object and make sure the file exists:-
File f = new File("title.txt");
if(f.exists() && !f.isDirectory()) {
fout = new FileOutputStream(f);
new PrintStream(fout).println(request.getParameter("txttitle"));
fout.close();
System.out.println(request.getParameter("txttitle"));
}
If the servlet cannot find the file give the full path to the file specified, like new File("D:\\Newfolder\\title.txt");

you should check first if the file doesn't exist ,create it
if(!new File("title.txt").exists())
{
File myfile = new File("title.txt");
myfile.createNewFile();
}
then you can use FileWriter or FileOutputStream to write to the file i prefer FileWriter
FileWriter writer = new FileWriter("title.txt");
writer.write("No God But Allah");
writer.close();
simply simple

How to copy files out of the currently running jar

I have a .jar that has two .dll files that it is dependent on. I would like to know if there is any way for me to copy these files from within the .jar to a users temp folder at runtime. here is the current code that I have (edited to just one .dll load to reduce question size):
public String tempDir = System.getProperty("java.io.tmpdir");
public String workingDir = dllInstall.class.getProtectionDomain().getCodeSource().getLocation().getPath();
public boolean installDLL() throws UnsupportedEncodingException {
try {
String decodedPath = URLDecoder.decode(workingDir, "UTF-8");
InputStream fileInStream = null;
OutputStream fileOutStream = null;
File fileIn = new File(decodedPath + "\\loadAtRuntime.dll");
File fileOut = new File(tempDir + "loadAtRuntime.dll");
fileInStream = new FileInputStream(fileIn);
fileOutStream = new FileOutputStream(fileOut);
byte[] bufferJNI = new byte[8192000013370000];
int lengthFileIn;
while ((lengthFileIn = fileInStream.read(bufferJNI)) > 0) {
fileOutStream.write(bufferJNI, 0, lengthFileIn);
}
//close all steams
} catch (IOException e) {
e.printStackTrace();
return false;
} catch (UnsupportedEncodingException e) {
System.out.println(e);
return false;
}
My main problem is getting the .dll files out of the jar at runtime. Any way to retrieve the path from within the .jar would be helpful.
Thanks in advance.

Since your dlls are bundeled inside your jar file you could just try to acasses them as resources using ClassLoader#getResourceAsStream and write them as binary files any where you want on the hard drive.
Here is some sample code:
InputStream ddlStream = <SomeClassInsideTheSameJar>.class
.getClassLoader().getResourceAsStream("some/pack/age/somelib.dll");
try (FileOutputStream fos = new FileOutputStream("somelib.dll");){
byte[] buf = new byte[2048];
int r;
while(-1 != (r = ddlStream.read(buf))) {
fos.write(buf, 0, r);
}
}
The code above will extract the dll located in the package some.pack.age to the current working directory.

Use a class loader that is able to locate resources in this JAR file. Either you can use the class loader of a class as Peter Lawrey suggested, or you can also create a URLClassLoader with the URL to that JAR.
Once you have that class loader you can retrieve a byte input stream with ClassLoader.getResourceAsStream. On the other hand you just create a FileOutputStream for the file you want to create.
The last step then is to copy all bytes from the input stream to the output stream, as you already did in your code example.

Use myClass.getClassLoader().getResourceAsStream("loadAtRuntime.dll"); and you will be able to find and copy DLLs in the JAR. You should pick a class which will also be in the same JAR.

Creating a file is causing problem, the File.getPath() doesn't seem to work

I am trying to create a back up file for an html file on a web server.
I want the backup to be in the same location as the existing file (it's a quick fix). I want to create the file using File file = new File(PathName);
public void backUpOldPage(String oldContent) throws IOException{
// this.uri is a class variable with the path of the file to be backed up
String fileName = new File(this.uri).getName();
String pathName = new File(this.uri).getPath();
System.out.println(pathName);
String bckPath = pathName+"\\"+bckName;
FileOutputStream fout;
try
{
// Open an output stream
fout = new FileOutputStream (bckFile);
fout.close();
}
// Catches any error conditions
catch (IOException e)
{
System.err.println ("Unable to write to file");
System.exit(-1);
}
}
But if instead I was to set bckPath like this, it will work.
String bckPath = "C://dev/server/tomcat6/webapps/sample-site/index_sdjf---sd.html";
I am working on Windows, not sure if that makes a difference.
The result of String bckPath = pathName+"\"+bckName;
is bckPath = C:\dev\server\tomcat6\webapps\sample-site\filename.html - this doesn't result in a new file.

Use File.pathSeparator, that way you dont need to worry what OS you are using.

Try to use File.getCanonicalPath() instead of plain getPath(). This helps if the orginal path is not fully specified.
Regarding slashes, / or \ or File.pathSeparator is not causing the problem, because they are all the same on Windows and Java. (And you do not define bckFile in your code, only bckPath. Also use getCanonicalPath() on the new created bckPath.)