Path FROM = Paths.get // need to get my file in my bin folder called s.txt, how would this be done?
Path TO = Paths.get("C:\\Temp\\to.txt");
try {
Files.copy(FROM, TO);
} catch (IOException e) {
e.printStackTrace();
}
Hi, I would really appreciate help, I basically need to get the path of a file located in my /bin/path
The local path to your project can be found using
System.getProperty("user.dir");
if your jar thats running is c:\workdir\myproject\bin\myproject.jar then System.getProperty("user.dir"); will return c:\workdir\myproject\bin
Heres how you could use it in your code...
Path FROM = Paths.get(System.getProperty("user.dir") + "/s.txt");
Path TO = Paths.get("C:\\Temp\\to.txt");
try {
Files.copy(FROM, TO);
} catch (IOException e) {
e.printStackTrace();
}
If I remember correctly, if using eclipse, go to:
project; properties; java build path; source; add folder - select bin folder
Then in your code do
Path TO = Paths.get(getClass().getResource("/bin/s.txt"));
Or you can defined the full system path from C:/ like your to path.
Related
Getting an error when trying to open a FileInputStream to load Map from file with .ser extension.
Constructor where I create new File and invoke method that loads map from file:
protected DriveatorImpl() {
accounts = new ConcurrentHashMap<String, Client>();
db = new File("database.ser"); // oddly this does not create a file if one does not exist
loadDB();
}
#SuppressWarnings("unchecked")
private void loadDB() {
try {
fileIn = new FileInputStream(db);
in = new ObjectInputStream(fileIn);
accounts = (Map<String, Client>) in.readObject();
in.close();
fileIn.close();
} catch (FileNotFoundException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
} catch (ClassNotFoundException e) {
e.printStackTrace();
}
}
I've tried to create file manually and put it in same package with class, but it does not help. What's going on?!
Thank You!
You provide a relative path for the file. That means program will look for the file relative to the working directory.
Depending on how you run the program it will be the directory you run it from (if run from Shell/Cmd) or whatever is configured in the project settings (if run from the IDE). For the latter, it depends on the IDE but usually it's the project root directory.
More info on working directory: https://en.wikipedia.org/wiki/Working_directory
More info on relative path: https://en.wikipedia.org/wiki/Path_(computing)#Absolute_and_relative_paths
Regarding creation of the file, it would create non-existing file if you were to write to it. When you read it, it expects it to exist. That means you have to create empty file (if one does not exist) before reading or simply treat exception as empty content.
The path to the file you have given might be wrong for IDE it can take relative path but from the command line, it will take the absolute path.
I've made this method that copies files from one absolute path (input directory) to another absolute path (output directory).
It doesn't give me any error, however no files are copied to the output folder.
Why would this be?
public static boolean copyFiles(String input, String output)
{
File source = new File(input);
File dest = new File(output);
try {
Files.copy(Paths.get(input), Paths.get(output), StandardCopyOption.REPLACE_EXISTING);
} catch (Exception e) {
e.printStackTrace();
return false;
}
return true;
}
As #zapl said, Files.copy() only copies the directory.
I found the solution, by importing the Apache commons.io library.
org.apache.commons.io.FileUtils.copyDirectory(new File(input), new File(output));
This works.
For my case, the Files are copied, just that it is not shown in the project explorer (in Eclipse) so just refresh it will do.
I am trying to load properties file. Here is my structure
Now i am trying to load test.properties file. But i am getting null. Here how i am doing
public class Test {
String workingDir = System.getProperty("user.dir");
System.out.println("Current working directory : " + workingDir);
File temp = new File(workingDir + "\\" + "test.properties");
String absolutePath = temp.getAbsolutePath();
System.out.println("File path : " + absolutePath);
Properties properties = null;
try {
properties = new Properties();
InputStream resourceAsStream = Test.class.getClassLoader().getResourceAsStream(absolutePath);
if (resourceAsStream != null) {
properties.load(resourceAsStream);
}
} catch (IOException e) {
e.printStackTrace();
}
System.exit(0);
} //end of class Test
This program prints
Current working directory : D:\Personal Work\eclipse 32 Bit\workspace\Spring Integration\LS360BatchImportIntegration
File path : D:\Personal Work\eclipse 32 Bit\workspace\Spring Integration\LS360BatchImportIntegration\test.properties
But it is not loading properties file from this path. Although it is present there. Why i am getting null ?
Thanks
Edit---
----------------------------
String workingDir = System.getProperty("user.dir");
System.out.println("Current working directory : " + workingDir);
File temp = new File(workingDir, "test.properties");
String absolutePath = temp.getAbsolutePath();
System.out.println("File path : " + absolutePath);
try {
properties = new Properties();
InputStream resourceAsStream = new FileInputStream(temp);
if (resourceAsStream != null) {
properties.load(resourceAsStream);
}
} catch (IOException e) {
e.printStackTrace();
}
System.exit(0);
Current working directory : D:\Personal Work\eclipse 32 Bit\workspace\Spring Integration\LS360BatchImportIntegration
File path : D:\Personal Work\eclipse 32 Bit\workspace\Spring Integration\LS360BatchImportIntegration\test.properties
java.io.FileNotFoundException: D:\Personal Work\eclipse 32 Bit\workspace\Spring Integration\LS360BatchImportIntegration\test.properties (The system cannot find the file specified)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(Unknown Source)
at com.softech.ls360.integration.BatchImport.main(BatchImport.java:57)
Oh oh ... There are several problems here:
1) In your first provided code snippet, you are using a ClassLoader for loading a resource file. This is indeed a good decision. But the getResourceAsStream method needs a "class-path relative" name. You are providing an absolute path.
2) Your second code snippet (after edit) results in not being able to find the file "D:...\LS360BatchImportIntegration\test.properties". According to your screenshot, the file should be "D:...\LS360AutomatedRegulatorsReportingService\test.properties". This is another directory.
I fear, that your descriptions are not up to date with the findings on your machine.
But let's just move to a reasonable solution:
1) In your Eclipse project (the screenshot tells us, that you are using Eclipse), create a new directory named "resources" in the same depth as your "src" directory. Copy - or better move - the properties file into it.
2) This new directory must be put into the "build path". Right-click the directory in the Package Explorer or Project Explorer view, select "Build Path", then "Use as Source Folder". Note: This build path will be the class path for the project, when you run it.
3) As the resources directory now is part of your class path and contains your properties file, you can simply load it with getResourceAsStream("test.properties").
EDIT
I just see, that you also use Maven (the pom.xml file). In Maven, such a resources directory exists by default and is part of the build path. It is "src/main/resources". If so, just use this.
Please put your property file in /src/main/resources folder and load from ClassLoader. It will be fix.
like
/src/main/resources/test.properties
Properties properties = null;
try {
properties = new Properties();
InputStream resourceAsStream = Test.class.getClassLoader().getResourceAsStream("test.properties");
if (resourceAsStream != null) {
properties.load(resourceAsStream);
}
} catch (IOException e) {
e.printStackTrace();
}
You are using the class loader (which reads in the classpath) whereas you are using the absolute path.
Simply try:
InputStream resourceAsStream = new FileInputStream(temp);
As a side note, try instanciating your file doing:
File temp = new File(workingDir, "test.properties");
to use the system-dependent path spearator.
I had a similar problem with a file not being found by getResourceAsStream(). The file was in the resources folder (src/main/resources), and still not found.
The problem got resolved when I went into the eclipse Package Explorer and "refreshed" the resources folder. It was in the directory, but Eclipse did not see it until the folder was refreshed (right-click on the folder and select Refresh).
I hope this helps !!
You can keep your test.properties into src/main/resources
public static Properties props = new Properties();
InputStream inStream = Test.class.getResourceAsStream("/test.properties");
try {
loadConfigurations(inStream);
} catch (IOException ex) {
String errMsg = "Exception in loading configuration file. Please check if application.properties file is present in classpath.";
ExceptionUtils.throwRuntimeException(errMsg, ex, LOGGER);
}
public static void loadConfigurations(InputStream inputStream) throws IOException{
props.load(inputStream);
}
You're passing a file path to getResourceAsStream(String name), but name here is a class path, not a file path...
You could make sure the file is on your classpath, or use a FileInputStream instead.
I have a resources folder/package in the root of my project, I "don't" want to load a certain File. If I wanted to load a certain File, I would use class.getResourceAsStream and I would be fine!! What I actually want to do is to load a "Folder" within the resources folder, loop on the Files inside that Folder and get a Stream to each file and read in the content... Assume that the File names are not determined before runtime... What should I do? Is there a way to get a list of the files inside a Folder in your jar File?
Notice that the Jar file with the resources is the same jar file from which the code is being run...
Finally, I found the solution:
final String path = "sample/folder";
final File jarFile = new File(getClass().getProtectionDomain().getCodeSource().getLocation().getPath());
if(jarFile.isFile()) { // Run with JAR file
final JarFile jar = new JarFile(jarFile);
final Enumeration<JarEntry> entries = jar.entries(); //gives ALL entries in jar
while(entries.hasMoreElements()) {
final String name = entries.nextElement().getName();
if (name.startsWith(path + "/")) { //filter according to the path
System.out.println(name);
}
}
jar.close();
} else { // Run with IDE
final URL url = Launcher.class.getResource("/" + path);
if (url != null) {
try {
final File apps = new File(url.toURI());
for (File app : apps.listFiles()) {
System.out.println(app);
}
} catch (URISyntaxException ex) {
// never happens
}
}
}
The second block just work when you run the application on IDE (not with jar file), You can remove it if you don't like that.
Try the following.
Make the resource path "<PathRelativeToThisClassFile>/<ResourceDirectory>" E.g. if your class path is com.abc.package.MyClass and your resoure files are within src/com/abc/package/resources/:
URL url = MyClass.class.getResource("resources/");
if (url == null) {
// error - missing folder
} else {
File dir = new File(url.toURI());
for (File nextFile : dir.listFiles()) {
// Do something with nextFile
}
}
You can also use
URL url = MyClass.class.getResource("/com/abc/package/resources/");
The following code returns the wanted "folder" as Path regardless of if it is inside a jar or not.
private Path getFolderPath() throws URISyntaxException, IOException {
URI uri = getClass().getClassLoader().getResource("folder").toURI();
if ("jar".equals(uri.getScheme())) {
FileSystem fileSystem = FileSystems.newFileSystem(uri, Collections.emptyMap(), null);
return fileSystem.getPath("path/to/folder/inside/jar");
} else {
return Paths.get(uri);
}
}
Requires java 7+.
I know this is many years ago . But just for other people come across this topic.
What you could do is to use getResourceAsStream() method with the directory path, and the input Stream will have all the files name from that dir. After that you can concat the dir path with each file name and call getResourceAsStream for each file in a loop.
I had the same problem at hands while i was attempting to load some hadoop configurations from resources packed in the jar... on both the IDE and on jar (release version).
I found java.nio.file.DirectoryStream to work the best to iterate over directory contents over both local filesystem and jar.
String fooFolder = "/foo/folder";
....
ClassLoader classLoader = foofClass.class.getClassLoader();
try {
uri = classLoader.getResource(fooFolder).toURI();
} catch (URISyntaxException e) {
throw new FooException(e.getMessage());
} catch (NullPointerException e){
throw new FooException(e.getMessage());
}
if(uri == null){
throw new FooException("something is wrong directory or files missing");
}
/** i want to know if i am inside the jar or working on the IDE*/
if(uri.getScheme().contains("jar")){
/** jar case */
try{
URL jar = FooClass.class.getProtectionDomain().getCodeSource().getLocation();
//jar.toString() begins with file:
//i want to trim it out...
Path jarFile = Paths.get(jar.toString().substring("file:".length()));
FileSystem fs = FileSystems.newFileSystem(jarFile, null);
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(fs.getPath(fooFolder));
for(Path p: directoryStream){
InputStream is = FooClass.class.getResourceAsStream(p.toString()) ;
performFooOverInputStream(is);
/** your logic here **/
}
}catch(IOException e) {
throw new FooException(e.getMessage());
}
}
else{
/** IDE case */
Path path = Paths.get(uri);
try {
DirectoryStream<Path> directoryStream = Files.newDirectoryStream(path);
for(Path p : directoryStream){
InputStream is = new FileInputStream(p.toFile());
performFooOverInputStream(is);
}
} catch (IOException _e) {
throw new FooException(_e.getMessage());
}
}
Another solution, you can do it using ResourceLoader like this:
import org.springframework.core.io.Resource;
import org.apache.commons.io.FileUtils;
#Autowire
private ResourceLoader resourceLoader;
...
Resource resource = resourceLoader.getResource("classpath:/path/to/you/dir");
File file = resource.getFile();
Iterator<File> fi = FileUtils.iterateFiles(file, null, true);
while(fi.hasNext()) {
load(fi.next())
}
If you are using Spring you can use org.springframework.core.io.support.PathMatchingResourcePatternResolver and deal with Resource objects rather than files. This works when running inside and outside of a Jar file.
PathMatchingResourcePatternResolver r = new PathMatchingResourcePatternResolver();
Resource[] resources = r.getResources("/myfolder/*");
Then you can access the data using getInputStream and the filename from getFilename.
Note that it will still fail if you try to use the getFile while running from a Jar.
As the other answers point out, once the resources are inside a jar file, things get really ugly. In our case, this solution:
https://stackoverflow.com/a/13227570/516188
works very well in the tests (since when the tests are run the code is not packed in a jar file), but doesn't work when the app actually runs normally. So what I've done is... I hardcode the list of the files in the app, but I have a test which reads the actual list from disk (can do it since that works in tests) and fails if the actual list doesn't match with the list the app returns.
That way I have simple code in my app (no tricks), and I'm sure I didn't forget to add a new entry in the list thanks to the test.
Below code gets .yaml files from a custom resource directory.
ClassLoader classLoader = this.getClass().getClassLoader();
URI uri = classLoader.getResource(directoryPath).toURI();
if("jar".equalsIgnoreCase(uri.getScheme())){
Pattern pattern = Pattern.compile("^.+" +"/classes/" + directoryPath + "/.+.yaml$");
log.debug("pattern {} ", pattern.pattern());
ApplicationHome home = new ApplicationHome(SomeApplication.class);
JarFile file = new JarFile(home.getSource());
Enumeration<JarEntry> jarEntries = file.entries() ;
while(jarEntries.hasMoreElements()){
JarEntry entry = jarEntries.nextElement();
Matcher matcher = pattern.matcher(entry.getName());
if(matcher.find()){
InputStream in =
file.getInputStream(entry);
//work on the stream
}
}
}else{
//When Spring boot application executed through Non-Jar strategy like through IDE or as a War.
String path = uri.getPath();
File[] files = new File(path).listFiles();
for(File file: files){
if(file != null){
try {
InputStream is = new FileInputStream(file);
//work on stream
} catch (Exception e) {
log.error("Exception while parsing file yaml file {} : {} " , file.getAbsolutePath(), e.getMessage());
}
}else{
log.warn("File Object is null while parsing yaml file");
}
}
}
Took me 2-3 days to get this working, in order to have the same url that work for both Jar or in local, the url (or path) needs to be a relative path from the repository root.
..meaning, the location of your file or folder from your src folder.
could be "/main/resources/your-folder/" or "/client/notes/somefile.md"
Whatever it is, in order for your JAR file to find it, the url must be a relative path from the repository root.
it must be "src/main/resources/your-folder/" or "src/client/notes/somefile.md"
Now you get the drill, and luckily for Intellij Idea users, you can get the correct path with a right-click on the folder or file -> copy Path/Reference.. -> Path From Repository Root (this is it)
Last, paste it and do your thing.
Simple ... use OSGi. In OSGi you can iterate over your Bundle's entries with findEntries and findPaths.
Inside my jar file I had a folder called Upload, this folder had three other text files inside it and I needed to have an exactly the same folder and files outside of the jar file, I used the code below:
URL inputUrl = getClass().getResource("/upload/blabla1.txt");
File dest1 = new File("upload/blabla1.txt");
FileUtils.copyURLToFile(inputUrl, dest1);
URL inputUrl2 = getClass().getResource("/upload/blabla2.txt");
File dest2 = new File("upload/blabla2.txt");
FileUtils.copyURLToFile(inputUrl2, dest2);
URL inputUrl3 = getClass().getResource("/upload/blabla3.txt");
File dest3 = new File("upload/Bblabla3.txt");
FileUtils.copyURLToFile(inputUrl3, dest3);
Using jdk7, I am trying to use the java.nio.file.Files class to move an empty directory, let's say Bar, into another empty directory, let's say Foo
Path source = Paths.get("Bar");
Path target = Paths.get("Foo");
try {
Files.move(
source,
target,
StandardCopyOption.REPLACE_EXISTING);
} catch (IOException e) {
e.printStackTrace();
}
After executing that code snippet, I expected that the Bar directory would be in the Foo directory (...\Foo\Bar). Instead it is not. And here's the kicker, it's been deleted as well. Also, no exceptions were thrown.
Am I doing this wrong?
NOTE
I'm looking for a jdk7-specific solution.I am also looking into the problem, but I figured I'd see if there was anyone else playing around with jdk7.
EDIT
In addition to the accepted answer, here's another solution
Path source = Paths.get("Bar");
Path target = Paths.get("Foo");
try {
Files.move(
source,
target.resolve(source.getFileName()),
StandardCopyOption.REPLACE_EXISTING);
} catch (IOException e) {
e.printStackTrace();
}
I didn't realize jdk7 java.nio.file.Files is a necessity, so here is the edited solution. Please see if it works coz I have never used the new Files class before.
Path source = Paths.get("Bar");
Path target = Paths.get("Foo", "Bar");
try {
Files.move(
source,
target,
StandardCopyOption.REPLACE_EXISTING);
} catch (IOException e) {
e.printStackTrace();
}
In the javadoc for the Files.move method you will find an example where it moves a file into a directory, keeping the same file name. This seems to be what you were looking for.
Here is the solution.
http://docs.oracle.com/javase/7/docs/api/java/nio/file/Files.html
suppose we want to move a file to new directory, keeping the same file name, and replacing any existing file of that name in the directory:
Path source = ...
Path newdir = ...
Files.move(source, newdir.resolve(source.getFileName()), REPLACE_EXISTING);
//Files.move(source, newdir.resolve(source.getFileName()), StandardCopyOption.REPLACE_EXISTING);