I am writing a camera application for the android platform. I am using the CameraKitView Library for producing my camera view. Everything else including accessing the camera is working as expected except for actually capturing and saving the image. minimum sdk is 15 and target sdk and compile sdk is 28. The code for saving the image is as shown below
cameraKitView.captureImage(new CameraKitView.ImageCallback() {
#Override
public void onImage(CameraKitView cameraKitView, byte[] photo) {
File savedPhoto = new File(Environment.getExternalStorageDirectory(), "pchk.jpg");
try{
FileOutputStream outputStream = new FileOutputStream(savedPhoto.getPath());
outputStream.write(photo);
outputStream.close();
}catch (java.io.IOException e){
e.printStackTrace();
}
}
});
Thank you in advance for your assistance
First of all verify that your file or folder where you want to save a file is exists or not, if not create them
capturedFolderPath is a path of the folder where you want to create the file
File folderFile = new File(capturedFolderPath)
if (!folderFile.exists()) {
folderFile.mkdirs();
}
String imageNameToSave = "xyz.jpg";
String imagePath = capturedFolderPath
+ File.separator + imageNameToSave + ".jpg";
File photoFile = new File(imagePath);
if (!photoFile.exists()) {
photoFile.createNewFile();
}
after creating write the bytes on the file like this
FileOutputStream outputStream = new FileOutputStream(photoFile);
outputStream.write(bytes);
outputStream.close();
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);
}
}
So, I need so clarity as to what I am overlooking when it comes to attempting to set a default image file path if a user selects no image. I utilized JFileChooser in a browse button in order to get the image path (relative path, not absolute) in order to store the image path on a MySQL database.
The problem is with my imageFileUploader class. I am attempting to set a generic image as a default if the user does not want to upload an image. When I test my imageFileUploader class, I am getting a null pointer exception. I essentially have the text box saying "no file uploaded" and if this stays true, then it uploads the default image. Can someone put an eye on my code and provide me some points to anything that I am overlooking? I would be much obliged.
String imagePath = "";
byte[] rawBytes = null;
FileInputStream fileInputStream = null;
if(imagePath.equals("No File Uploaded")) {//null pointer exeception here!!
ClassLoader classLoader = this.getClass().getClassLoader();
URL resources = classLoader.getResource("/abaApplicationImages/blankProfile.jpg");
imagePath = resources.getFile();
}else{
//File fileObject = new File(imagePath);
File fileObject = new File(imagePath);
try {
fileInputStream = new FileInputStream(fileObject);
} catch (FileNotFoundException e) {
e.printStackTrace();
System.out.println("file not found");
}
int imageLength = Integer.parseInt(String.valueOf(fileObject.length()));
rawBytes = new byte[imageLength];
try {
fileInputStream.read(rawBytes, 0, imageLength);
} catch (IOException e) {
e.printStackTrace();
System.out.println("IO exception");
}
}
return imagePath;
}
Error message:
java.io.FileNotFoundException: at java.io.FileInputStream.open(Native
Method) at java.io.FileInputStream.(Unknown Source) at
abaApplication.ImageUploader.ImageUploader(ImageUploader.java:30) at
abaApplication.ImageUploader.main(ImageUploader.java:59) file not
found Exception in thread "main" java.lang.NullPointerException at
abaApplication.ImageUploader.ImageUploader(ImageUploader.java:40) at
abaApplication.ImageUploader.main(ImageUploader.java:59)
You're not verifying if the URL resources is null or not. That's probably where the NPE is coming from.
Also ClassLoader.getResource() loads resource from the classpath, if the file is on the image file is not on classpath then you should probably use FileInputStream directly.
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.
I am trying to get a path to a Resource but I have had no luck.
This works (both in IDE and with the JAR) but this way I can't get a path to a file, only the file contents:
ClassLoader classLoader = getClass().getClassLoader();
PrintInputStream(classLoader.getResourceAsStream("config/netclient.p"));
If I do this:
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource("config/netclient.p").getFile());
The result is:
java.io.FileNotFoundException: file:/path/to/jarfile/bot.jar!/config/netclient.p (No such file or directory)
Is there a way to get a path to a resource file?
This is deliberate. The contents of the "file" may not be available as a file. Remember you are dealing with classes and resources that may be part of a JAR file or other kind of resource. The classloader does not have to provide a file handle to the resource, for example the jar file may not have been expanded into individual files in the file system.
Anything you can do by getting a java.io.File could be done by copying the stream out into a temporary file and doing the same, if a java.io.File is absolutely necessary.
When loading a resource make sure you notice the difference between:
getClass().getClassLoader().getResource("com/myorg/foo.jpg") //relative path
and
getClass().getResource("/com/myorg/foo.jpg")); //note the slash at the beginning
I guess, this confusion is causing most of problems when loading a resource.
Also, when you're loading an image it's easier to use getResourceAsStream():
BufferedImage image = ImageIO.read(getClass().getResourceAsStream("/com/myorg/foo.jpg"));
When you really have to load a (non-image) file from a JAR archive, you might try this:
File file = null;
String resource = "/com/myorg/foo.xml";
URL res = getClass().getResource(resource);
if (res.getProtocol().equals("jar")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile("tempfile", ".tmp");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.close();
file.deleteOnExit();
} catch (IOException ex) {
Exceptions.printStackTrace(ex);
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}
if (file != null && !file.exists()) {
throw new RuntimeException("Error: File " + file + " not found!");
}
The one line answer is -
String path = this.getClass().getClassLoader().getResource(<resourceFileName>).toExternalForm()
Basically getResource method gives the URL. From this URL you can extract the path by calling toExternalForm().
I spent a while messing around with this problem, because no solution I found actually worked, strangely enough! The working directory is frequently not the directory of the JAR, especially if a JAR (or any program, for that matter) is run from the Start Menu under Windows. So here is what I did, and it works for .class files run from outside a JAR just as well as it works for a JAR. (I only tested it under Windows 7.)
try {
//Attempt to get the path of the actual JAR file, because the working directory is frequently not where the file is.
//Example: file:/D:/all/Java/TitanWaterworks/TitanWaterworks-en.jar!/TitanWaterworks.class
//Another example: /D:/all/Java/TitanWaterworks/TitanWaterworks.class
PROGRAM_DIRECTORY = getClass().getClassLoader().getResource("TitanWaterworks.class").getPath(); // Gets the path of the class or jar.
//Find the last ! and cut it off at that location. If this isn't being run from a jar, there is no !, so it'll cause an exception, which is fine.
try {
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('!'));
} catch (Exception e) { }
//Find the last / and cut it off at that location.
PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(0, PROGRAM_DIRECTORY.lastIndexOf('/') + 1);
//If it starts with /, cut it off.
if (PROGRAM_DIRECTORY.startsWith("/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(1, PROGRAM_DIRECTORY.length());
//If it starts with file:/, cut that off, too.
if (PROGRAM_DIRECTORY.startsWith("file:/")) PROGRAM_DIRECTORY = PROGRAM_DIRECTORY.substring(6, PROGRAM_DIRECTORY.length());
} catch (Exception e) {
PROGRAM_DIRECTORY = ""; //Current working directory instead.
}
This is same code from user Tombart with stream flush and close to avoid incomplete temporary file content copy from jar resource and to have unique temp file names.
File file = null;
String resource = "/view/Trial_main.html" ;
URL res = getClass().getResource(resource);
if (res.toString().startsWith("jar:")) {
try {
InputStream input = getClass().getResourceAsStream(resource);
file = File.createTempFile(new Date().getTime()+"", ".html");
OutputStream out = new FileOutputStream(file);
int read;
byte[] bytes = new byte[1024];
while ((read = input.read(bytes)) != -1) {
out.write(bytes, 0, read);
}
out.flush();
out.close();
input.close();
file.deleteOnExit();
} catch (IOException ex) {
ex.printStackTrace();
}
} else {
//this will probably work in your IDE, but not from a JAR
file = new File(res.getFile());
}
if netclient.p is inside a JAR file, it won't have a path because that file is located inside other file. in that case, the best path you can have is really file:/path/to/jarfile/bot.jar!/config/netclient.p.
You need to understand the path within the jar file.
Simply refer to it relative. So if you have a file (myfile.txt), located in foo.jar under the \src\main\resources directory (maven style). You would refer to it like:
src/main/resources/myfile.txt
If you dump your jar using jar -tvf myjar.jar you will see the output and the relative path within the jar file, and use the FORWARD SLASHES.
In my case, I have used a URL object instead Path.
File
File file = new File("my_path");
URL url = file.toURI().toURL();
Resource in classpath using classloader
URL url = MyClass.class.getClassLoader().getResource("resource_name")
When I need to read the content, I can use the following code:
InputStream stream = url.openStream();
And you can access the content using an InputStream.
follow code!
/src/main/resources/file
streamToFile(getClass().getClassLoader().getResourceAsStream("file"))
public static File streamToFile(InputStream in) {
if (in == null) {
return null;
}
try {
File f = File.createTempFile(String.valueOf(in.hashCode()), ".tmp");
f.deleteOnExit();
FileOutputStream out = new FileOutputStream(f);
byte[] buffer = new byte[1024];
int bytesRead;
while ((bytesRead = in.read(buffer)) != -1) {
out.write(buffer, 0, bytesRead);
}
return f;
} catch (IOException e) {
LOGGER.error(e.getMessage(), e);
return null;
}
}
A File is an abstraction for a file in a filesystem, and the filesystems don't know anything about what are the contents of a JAR.
Try with an URI, I think there's a jar:// protocol that might be useful for your purpouses.
The following path worked for me: classpath:/path/to/resource/in/jar
private static final String FILE_LOCATION = "com/input/file/somefile.txt";
//Method Body
InputStream invalidCharacterInputStream = URLClassLoader.getSystemResourceAsStream(FILE_LOCATION);
Getting this from getSystemResourceAsStream is the best option. By getting the inputstream rather than the file or the URL, works in a JAR file and as stand alone.
It may be a little late but you may use my library KResourceLoader to get a resource from your jar:
File resource = getResource("file.txt")
Inside your ressources folder (java/main/resources) of your jar add your file (we assume that you have added an xml file named imports.xml), after that you inject ResourceLoader if you use spring like bellow
#Autowired
private ResourceLoader resourceLoader;
inside tour function write the bellow code in order to load file:
Resource resource = resourceLoader.getResource("classpath:imports.xml");
try{
File file;
file = resource.getFile();//will load the file
...
}catch(IOException e){e.printStackTrace();}
Maybe this method can be used for quick solution.
public class TestUtility
{
public static File getInternalResource(String relativePath)
{
File resourceFile = null;
URL location = TestUtility.class.getProtectionDomain().getCodeSource().getLocation();
String codeLocation = location.toString();
try{
if (codeLocation.endsWith(".jar"){
//Call from jar
Path path = Paths.get(location.toURI()).resolve("../classes/" + relativePath).normalize();
resourceFile = path.toFile();
}else{
//Call from IDE
resourceFile = new File(TestUtility.class.getClassLoader().getResource(relativePath).getPath());
}
}catch(URISyntaxException ex){
ex.printStackTrace();
}
return resourceFile;
}
}
When in a jar file, the resource is located absolutely in the package hierarchy (not file system hierarchy). So if you have class com.example.Sweet loading a resource named ./default.conf then the resource's name is specified as /com/example/default.conf.
But if it's in a jar then it's not a File ...
This Class function can get absolute file path by relative file path in .jar file.
public class Utility {
public static void main(String[] args) throws Exception {
Utility utility = new Utility();
String absolutePath = utility.getAbsolutePath("./absolute/path/to/file");
}
public String getAbsolutePath(String relativeFilePath) throws IOException {
URL url = this.getClass().getResource(relativeFilePath);
return url.getPath();
}
}
If target file is same directory with Utility.java, your input will be ./file.txt.
When you input only / on getAbsolutePath(), it returns /Users/user/PROJECT_NAME/target/classes/. This means you can select the file like this /com/example/file.txt.