I have the following code to insert "rose.gif" into a roseNode. But how do I retrieve the file from the repository?
Node roseNode = session.getRootNode().getNode("wiki:encyclopedia/wiki:entry[1]/");
File file = new File("rose.gif");
MimeTable mt = MimeTable.getDefaultTable();
String mimeType = mt.getContentTypeFor(file.getName());
if (mimeType == null) mimeType = "application/octet-stream";
Node fileNode = roseNode.addNode(file.getName(), "nt:file");
System.out.println( fileNode.getName() );
Node resNode = fileNode.addNode("jcr:content", "nt:resource");
resNode.setProperty("jcr:mimeType", mimeType);
resNode.setProperty("jcr:encoding", "");
resNode.setProperty("jcr:data", new FileInputStream(file));
Calendar lastModified = Calendar.getInstance();
lastModified.setTimeInMillis(file.lastModified());
resNode.setProperty("jcr:lastModified", lastModified);
//retrieve file and output as rose-out.gif
File outputFile = new File("rose-out.gif");
FileOutputStream out = new FileOutputStream(outputFile);
The only thing you really need to do is get the name of the file from the name of the "nt:file" node, and the content for the file from the "jcr:data" property on the "jcr:content" child node.
JCR 1.0 and 2.0 differ a bit in how you get the stream for the binary "jcr:data" property value. If you're using JCR 1.0, then the code would be like this:
Node fileNode = // find this somehow
Node jcrContent = fileNode.getNode("jcr:content");
String fileName = fileNode.getName();
InputStream content = jcrContent.getProperty("jcr:data").getStream();
If you're using JCR 2.0, the last line is a bit different because you first have to get the Binary object from the property value:
InputStream content = jcrContent.getProperty("jcr:data").getBinary().getStream();
You can then use standard Java stream utility to write the bytes from the 'content' stream into the file.
When you're done with the Binary object, be sure to call the Binary's dispose() method to tell signal that you're done with the Binary and that the implementation can release all resources acquired by the Binary object. You should always do this, even though some JCR implementations try to catch programming errors by returning a stream that, when closed, will automatically call dispose() for you.
Related
To merge Storage files in Codename One I elaborated this solution:
/**
* Merges the given list of Storage files in the output Storage file.
* #param toBeMerged
* #param output
* #throws IOException
*/
public static synchronized void mergeStorageFiles(List<String> toBeMerged, String output) throws IOException {
if (toBeMerged.contains(output)) {
throw new IllegalArgumentException("The output file cannot be contained in the toBeMerged list of input files.");
}
// Note: the temporary file used for merging is placed in the FileSystemStorage because it offers the method
// openOutputStream(String file, int offset) that allows appending to a stream. Storage doesn't have a such method.
long writtenBytes = 0;
String tempFile = FileSystemStorage.getInstance().getAppHomePath() + "/tempFileUsedInMerge";
for (String partialFile : toBeMerged) {
InputStream in = Storage.getInstance().createInputStream(partialFile);
OutputStream out = FileSystemStorage.getInstance().openOutputStream(tempFile, (int) writtenBytes);
Util.copy(in, out);
writtenBytes = FileSystemStorage.getInstance().getLength(tempFile);
}
Util.copy(FileSystemStorage.getInstance().openInputStream(tempFile), Storage.getInstance().createOutputStream(output));
FileSystemStorage.getInstance().delete(tempFile);
}
This solution is based on the API FileSystemStorage.openOutputStream(String file, int offset), that is the only API that I found to allow to append the content of a file to another.
Are there other API that can be used to append or merge files?
Thank you
Since you end up copying everything to a Storage entry I don't see the value of using FileSystemStorage as an intermediate merging tool.
The only reason I can think of is integrity of the output file (e.g. if failure happens while writing) but that can happen here too. You can guarantee integrity by setting a flag e.g. creating a file called "writeLock" and deleting it when write has finished successfully.
To be clear I would copy like this which is simpler/faster:
try(OutputStream out = Storage.getInstance().createOutputStream(output)) {
for (String partialFile : toBeMerged) {
try(InputStream in = Storage.getInstance().createInputStream(partialFile)) {
Util.copyNoClose(in, out, 8192);
}
}
}
I have several files with datas in it.
For example: file01.csv with x lignes in it, file02.csv with y lines in it.
I would like to treat and merge them with mapreduce in order to get a file with the x lines beginning with file01 then line content, and y files beginning with file02 then line content.
I have two issues here:
I know how to get lines from a file with mapreduce by setting FileInputFormat.setInputPath(job, new Path(inputFile));
But I don't understand how I can get lines of each file of a folder.
Once I have those lines in my mapper, how can I access to the filename corresponding, so that I can create the data I want ?
Thank you for your consideration.
Ambre
You do not need map-reduce in your situation. That's because you want to preserve the order of lines in result file. In this case single thread processing will be faster.
Just run java client with code like this:
FileSystem fs = FileSystem.get();
OutputStream os = fs.create(outputPath); // stream for result file
PrintWriter pw = new PrintWriter(new OutputStreamWriter(os));
for (String inputFile : inputs) { // reading input files
InputStream is = fs.open(new Path(inputFile));
BufferedReader br = new BufferedReader(new InputStreamReader(is));
String line;
while ((line = br.readLine()) != null) {
pw.println(line);
}
br.close();
}
pw.close();
I am writing an Eclipse Plugin which requires me to get full path of any kind of file open in the Workspace.
I am able to get full path of any file which is part of any Eclipse project. Code to get open/active editor file from workspace.
public static String getActiveFilename(IWorkbenchWindow window) {
IWorkbenchPage activePage = window.getActivePage();
IEditorInput input = activePage.getActiveEditor().getEditorInput();
String name = activePage.getActiveEditor().getEditorInput().getName();
PluginUtils.log(activePage.getActiveEditor().getClass() +" Editor.");
IPath path = input instanceof FileEditorInput ? ((FileEditorInput) input).getPath() : null;
if (path != null) {
return path.toPortableString();
}
return name;
}
However, if any file is drag-dropped in Workspace or opened using File -> Open File. For instance, I opened a file from /Users/mac/log.txt from File -> Open File. My plugin is not able to find location of this file.
After couple of days search, I found the answer by looking at the source code of Eclipse IDE.
In IDE.class, Eclipse tries to find a suitable editor input depending on the workspace file or an external file. Eclipse handles files in workspace using FileEditorInput and external files using FileStoreEditorInput. Code snippet below:
/**
* Create the Editor Input appropriate for the given <code>IFileStore</code>.
* The result is a normal file editor input if the file exists in the
* workspace and, if not, we create a wrapper capable of managing an
* 'external' file using its <code>IFileStore</code>.
*
* #param fileStore
* The file store to provide the editor input for
* #return The editor input associated with the given file store
* #since 3.3
*/
private static IEditorInput getEditorInput(IFileStore fileStore) {
IFile workspaceFile = getWorkspaceFile(fileStore);
if (workspaceFile != null)
return new FileEditorInput(workspaceFile);
return new FileStoreEditorInput(fileStore);
}
I have modified the code posted in the question to handle both files in Workspace and external file.
public static String getActiveEditorFilepath(IWorkbenchWindow window) {
IWorkbenchPage activePage = window.getActivePage();
IEditorInput input = activePage.getActiveEditor().getEditorInput();
String name = activePage.getActiveEditor().getEditorInput().getName();
//Path of files in the workspace.
IPath path = input instanceof FileEditorInput ? ((FileEditorInput) input).getPath() : null;
if (path != null) {
return path.toPortableString();
}
//Path of the externally opened files in Editor context.
try {
URI urlPath = input instanceof FileStoreEditorInput ? ((FileStoreEditorInput) input).getURI() : null;
if (urlPath != null) {
return new File(urlPath.toURL().getPath()).getAbsolutePath();
}
} catch (MalformedURLException e) {
e.printStackTrace();
}
//Fallback option to get at least name
return name;
}
In my app I tried to pass the file path from one activity to another activity using intent.In my receiving activity I got the file path as "null".But when I print the file in first activity it prints the path.From my second activity I attach that file to mail using Gmailsender.This was the code I tried,
private void startRecord()
{
File file = new File(Environment.getExternalStorageDirectory(), "test.pcm");
try
{
file.createNewFile();
OutputStream outputStream = new FileOutputStream(file);
BufferedOutputStream bufferedOutputStream = new BufferedOutputStream(outputStream);
DataOutputStream dataOutputStream = new DataOutputStream(bufferedOutputStream);
int minBufferSize = AudioRecord.getMinBufferSize(8000,
AudioFormat.CHANNEL_IN_MONO,
AudioFormat.ENCODING_PCM_16BIT);
short[] audioData = new short[minBufferSize];
AudioRecord audioRecord = new AudioRecord(MediaRecorder.AudioSource.MIC,
8000,
AudioFormat.CHANNEL_IN_MONO,
AudioFormat.ENCODING_PCM_16BIT,
minBufferSize);
audioRecord.startRecording();
while(recording)
{
int numberOfShort = audioRecord.read(audioData, 0, minBufferSize);
for(int i = 0; i < numberOfShort; i++)
{
dataOutputStream.writeShort(audioData[i]);
}
}
audioRecord.stop();
audioRecord.release();
dataOutputStream.close();
}
catch (IOException e)
{
e.printStackTrace();
}
String audiofile;
audiofile=file.getAbsolutePath();
System.out.println("File Path::::"+audiofile);
}
Intent is,
Intent sigout=new Intent(getApplicationContext(),WeeklyendActivity.class);
sigout.putExtra("mnt/sdcard-test.pcm",audiofile);
startActivity(sigout);
In my receiving activity,
String patty=getIntent().getStringExtra("mnt/sdcard-text.pcm");
System.out.println("paathhhy frfom ::"+patty);
It prints null.Can anyone help me how to get the file path.And more thing I am not sure whether the audio would save in that file correctly?
Please anyone help me!!!Thanks in advance!
Based on your information that audioFile is a variable of type File, when you do this:
sigout.putExtra("mnt/sdcard-test.pcm",audiofile);
you are putting a File object in the extras Bundle. Then, when you try to get the extra from the Bundle you do this:
String patty=getIntent().getStringExtra("mnt/sdcard-text.pcm");
However, the object in this extra is of type File, not type String. This is why you are getting null.
If you only want to pass the name of the file, then put the extra like this:
sigout.putExtra("mnt/sdcard-test.pcm",audiofile.getAbsolutePath());
/*
*This program checks type casting from String to int/double from a file
*/
import java.io.*;
import java.lang.String;
public class ConvertingStringsToNums {
public static void main (String[] args){
File dataFile = new File("/files/scores.dat");
FileReader in;
BufferedReader readFile;
String score;
double avgScore, totalScores = 0;
int numScores = 0;
//------------------------------------------------------------
try {
in = new FileReader(dataFile);
readFile = new BufferedReader(in);
while((score = readFile.readLine()) != null) {
numScores += 1;
System.out.println(score);
totalScores += Double.parseDouble(score);
}
avgScore = totalScores / numScores;
readFile.close();
in.close();
} catch(FileNotFoundException e) {
System.err.println("FileNotFoundException: " + e.getMessage());
} catch (IOException e) {
System.err.println("IOException: " + e.getMessage());
} //end try/catch
}
}
1) If you wish to open a file at an absolute file path on your hard drive:
br = new BufferedReader (
new FileReader(
new File ("/files/scores.dat")));
2) If you wish to open a file at an relative path relative to where you started your app:
br = new BufferedReader (
new FileReader(
new File ("files/scores.dat")));
3) If you wish to open a file at an relative path relative to your class files (particularly relevant for packages and/or for executing from a .jar or a .war):
this.getClass().getResourceAsStream ("files/scores.dat");
'Hope that helps
The reason is can be that you wont be having permission to open the file.
try chmod 755 scores.dat from terminal in order to change the permissions and see if the error still exist.
The answer to this problem exists in the javadocs for the File class:
For UNIX platforms, the prefix of an absolute pathname is always "/". Relative pathnames have no prefix. The abstract pathname denoting the root directory has the prefix "/" and an empty name sequence.
In your code, you have the following:
File dataFile = new File("/files/scores.dat");
According to the documentation, this is an absolute path, which means Java is looking for a folder at the root of the filesystem called "files" and then looking for scores.dat in that folder.
If you instead expect to search for a files directory that is relative to the current directory, you'd need to omit the first /:
File dataFile = new File("files/scores.dat");
The other option is to use an absolute path to your data file, but you may run into problems if you change the location of your project or put the class files in a JAR file.
Try turning up your logging level to DEBUG or ALL so that you can see exactly where the program is trying to look. This will help you adjust your code to target the right folder.