When I run this as a jar, this .properties file is created on the desktop. For the sake of keeping things clean, how can I set the path to save this file somewhere else, like a folder? Or even the jar itself but I was having trouble getting that to work. I plan on giving this to someone and wouldn't want their desktop cluttered in .properties files..
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.OutputStream;
import java.util.Properties;
public class DataFile {
public static void main(String[] args) {
Properties prop = new Properties();
OutputStream output = null;
try {
output = new FileOutputStream("config.properties");
prop.setProperty("prop1", "000");
prop.setProperty("prop2", "000");
prop.setProperty("prop3", "000");
prop.store(output, null);
} catch (IOException io) {
io.printStackTrace();
} finally {
if (output != null) {
try {
output.close();
} catch (IOException e) {
e.printStackTrace();
}
}
}
}
}
Since you are using the file name without a path, the file you creates ends in the CWD. That is the Current working directory the process inherits from the OS.
That is is you execute your jar file from the Desktop directory any files that use relative paths will end in the Desktop or any of it sub directories.
To control the absolute location of the file you must use absolute paths.
An absolute path always starts with a slash '/'.
Absolute path:
/etc/config.properties
Relative path:
sub_dir/config.properties
The simplest way is to hard code some path into the file path string.
output = new FileOutputStream("/etc/config.properties");
You can of course setup the path in a property which you can pass using the command line instead of hard coding it. The you concat the path name and the file name together.
String path = "/etc";
String full_path = "/etc" + "/" + "config.properties";
output = new FileOutputStream(full_path);
Please note that windows paths in java use a forward slash instead of back slash.
Check this for more details
file path Windows format to java format
Related
Can someone please point out what I'm doing wrong here.
I have a small weather app that generates and sends an HTML email. With my code below, everything works fine when I run it from Eclipse. My email gets generated, it's able to access my image resources and it sends the email with the included attachment.
However, when I build the executable jar by running mvn install and run the jar using java -jar NameOfMyJar.jar I get java.io.FileNotFound Exceptions for my image resource.
I know that I have to be doing something wrong with how I'm accessing my image resources, I just don't understand why it works fine when it's not packaged, but bombs out whenever I package it into a jar.
Any advice is very much appreciated it.
My project layout
How I'm accessing my image resource
//Setup the ATTACHMENTS
MimeBodyPart attachmentsPart = new MimeBodyPart();
try {
attachmentsPart.attachFile("resources/Cloudy_Day.png");
} catch (IOException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
The StackTrace
Exception in thread "main" java.lang.RuntimeException: javax.mail.MessagingException: IOException while sending message;
nested exception is:
java.io.FileNotFoundException: resources/Cloudy_Day.png (No such file or directory)
at Utilities.SendEmailUsingGmailSMTP.SendTheEmail(SendEmailUsingGmailSMTP.java:139)
at Utilities.SendEmailUsingGmailSMTP.SendWeatherEmail(SendEmailUsingGmailSMTP.java:66)
at Weather.Main.start(Main.java:43)
at Weather.Main.main(Main.java:23)
Caused by: javax.mail.MessagingException: IOException while sending message;
nested exception is:
java.io.FileNotFoundException: resources/Cloudy_Day.png (No such file or directory)
at com.sun.mail.smtp.SMTPTransport.sendMessage(SMTPTransport.java:1167)
at javax.mail.Transport.send0(Transport.java:195)
at javax.mail.Transport.send(Transport.java:124)
at Utilities.SendEmailUsingGmailSMTP.SendTheEmail(SendEmailUsingGmailSMTP.java:134)
... 3 more
Caused by: java.io.FileNotFoundException: resources/Cloudy_Day.png (No such file or directory)
at java.io.FileInputStream.open(Native Method)
at java.io.FileInputStream.<init>(FileInputStream.java:146)
at javax.activation.FileDataSource.getInputStream(FileDataSource.java:97)
at javax.activation.DataHandler.writeTo(DataHandler.java:305)
at javax.mail.internet.MimeBodyPart.writeTo(MimeBodyPart.java:1485)
at javax.mail.internet.MimeBodyPart.writeTo(MimeBodyPart.java:865)
at javax.mail.internet.MimeMultipart.writeTo(MimeMultipart.java:462)
at com.sun.mail.handlers.multipart_mixed.writeTo(multipart_mixed.java:103)
at javax.activation.ObjectDataContentHandler.writeTo(DataHandler.java:889)
at javax.activation.DataHandler.writeTo(DataHandler.java:317)
at javax.mail.internet.MimeBodyPart.writeTo(MimeBodyPart.java:1485)
at javax.mail.internet.MimeMessage.writeTo(MimeMessage.java:1773)
at com.sun.mail.smtp.SMTPTransport.sendMessage(SMTPTransport.java:1119)
... 6 more
Others are correct with the use of getResourceAsStream, but the path is a little tricky. You see the little package icon in the resources folder? That signifies that all the files in the resource folder will be put into the root of the classpath. Just like all the packages in src/main/java are put in the root. So you would take out the resources from the path
InputStream is = getClass().getResourceAsStream("/Cloudy_Day.png");
An aside: Maven has a file structure conventions. Class path resources are usually put into src/main/resources. If you create a resources dir in the src/main, Eclipse should automatically pick it up, and create the little package icon for a path src/main/resource that you should see in the project explorer. These files would also go to the root and could be accessed the same way. I would fix the file structure to follow this convention.
Note: A MimeBodyPart, can be Constructed from an InputStream (As suggested by Bill Shannon, this is incorrect). As mentioned in his comment below
"You can also attach the data using"
mbp.setDataHandler(new DataHandler(new ByteArrayDataSource(
this.getClass().getResourceAsStream("/Cloudy_Day.png", "image/png"))));
You can't access resources inside a JAR file as a File, only read them as an InputStream: getResourceAsStream().
As the MimeBodyPart has no attach() method for an InputStream, the easiest way should be to read your resources and write them to temp files, then attach these files.
Try this
new MimeBodyPart().attachFile(new File(this.getClass().getClassLoader().getResource("resources/Cloudy_Day.png").toURI());
I don't know if this will help anyone or not. But, I have a similar case as the OP and I solved the case by finding the file in the classpath using recursive function. The idea is so that when another developer decided to move the resources into another folder/path. It will still be found as long as the name is still the same.
For example, in my work we usually put our resources outside the jar, and then we add said resources path into our classpath, so here the classpath of the resources will be different depending on where it is located.
That's where my code comes to work, no matter where the file is put, as long as it's in the classpath it will be found.
Here is an example of my code in action:
import java.io.File;
public class FindResourcesRecursive {
public File findConfigFile(String paths, String configFilename) {
for (String p : paths.split(File.pathSeparator)) {
File result = findConfigFile(new File(p), configFilename);
if (result != null) {
return result;
}
}
return null;
}
private File findConfigFile(File path, String configFilename) {
if (path.isDirectory()) {
String[] subPaths = path.list();
if (subPaths == null) {
return null;
}
for (String sp : subPaths) {
File subPath = new File(path.getAbsoluteFile() + "/" + sp);
File result = findConfigFile(subPath, configFilename);
if (result != null && result.getName().equalsIgnoreCase(configFilename)) {
return result;
}
}
return null;
} else {
File file = path;
if (file.getName().equalsIgnoreCase(configFilename)) {
return file;
}
return null;
}
}
}
Here I have a test case that is coupled with a file "test.txt" in my test/resources folder. The content of said file is:
A sample file
Now, here is my test case:
import org.junit.Test;
import java.io.*;
import static org.junit.Assert.fail;
public class FindResourcesRecursiveTest {
#Test
public void testFindFile() {
// Here in the test resources I have a file "test.txt"
// Inside it is a string "A sample file"
// My Unit Test will use the class FindResourcesRecursive to find the file and print out the results.
File testFile = new FindResourcesRecursive().findConfigFile(
System.getProperty("java.class.path"),
"test.txt"
);
try (FileInputStream is = new FileInputStream(testFile)) {
int i;
while ((i = is.read()) != -1) {
System.out.print((char) i);
}
System.out.println();
} catch (IOException e) {
fail();
}
}
}
Now, if you run this test, it will print out "A sample file" and the test will be green.
I am trying to createNewFile() in java.I have written down the following example.I have compiled it but am getting a run time error.
import java.io.File;
import java.io.IOException;
public class CreateFileExample
{
public static void main(String [] args)
{
try
{
File file = new File("home/karthik/newfile.txt");
if(file.createNewFile())
{
System.out.println("created new fle");
}else
{
System.out.println("could not create a new file");
}
}catch(IOException e )
{
e.printStackTrace();
}
}
}
It is compiling OK.The run time error that I am getting is
java.io.IOException: No such file or directory
at java.io.UnixFileSystem.createFileExclusively(Native Method)
at java.io.File.createNewFile(File.java:947)
at CreateFileExample.main(CreateFileExample.java:16)
some points here
1- as Victor said you are missing the leading slash
2- if your file is created, then every time you invoke this method "File.createNewFile()" will return false
3- your class is very platform dependent (one of the main reasons why Java is powerful programming language is that it is a NON-PLATFORM dependent), instead you can detect a relative location throw using the System.getProperties() :
// get System properties :
java.util.Properties properties = System.getProperties();
// to print all the keys in the properties map <for testing>
properties.list(System.out);
// get Operating System home directory
String home = properties.get("user.home").toString();
// get Operating System separator
String separator = properties.get("file.separator").toString();
// your directory name
String directoryName = "karthik";
// your file name
String fileName = "newfile.txt";
// create your directory Object (wont harm if it is already there ...
// just an additional object on the heap that will cost you some bytes
File dir = new File(home+separator+directoryName);
// create a new directory, will do nothing if directory exists
dir.mkdir();
// create your file Object
File file = new File(dir,fileName);
// the rest of your code
try {
if (file.createNewFile()) {
System.out.println("created new fle");
} else {
System.out.println("could not create a new file");
}
} catch (IOException e) {
e.printStackTrace();
}
this way you will create your file in any home directory on any platform, this worked for my windows operating system, and is expected to work for your Linux or Ubuntu as well
You're missing the leading slash in the file path.
Try this:
File file = new File("/home/karthik/newfile.txt");
That should work!
Actually this error comes when there is no directory "karthik" as in above example and createNewFile() is only to create file not for directory use mkdir() for directory and then createNewFile() for file.
I am trying to read a file in Java called "KFormLList.txt". It is saved in the default package along with this program, yet when I try to run it I get this error message: "Error: KFormLList.txt (The system cannot find the file specified)"
What am I doing wrong? Thanks for any help.
import java.io.*;
public class VLOCGenerater {
/**
* #param args
*/
public static void main(String[] args) {
try {
//Read the text file "KFormLList.txt"
FileInputStream fis = new FileInputStream("KFormLList.txt");
DataInputStream dis = new DataInputStream(fis);
BufferedReader br = new BufferedReader(new InputStreamReader(dis));
String strLine;
int V = 0;
int LOC = 0;
while((strLine = br.readLine())!= null ){
if (strLine.trim().length() != 0){
System.out.println(strLine);
V++;
}
else {
LOC++;
}
}
System.out.println("V = " + V);
System.out.println("LOC = " + LOC);
dis.close();
}
catch (Exception e){
System.err.println("Error: " + e.getMessage());
}
}
}
Try putting the text file in the root directory of your project.
The name parameter passed into the FileInputStream constructor is the path name to the file in the file system.
A pathname, whether abstract or in string form, may be either absolute
or relative. An absolute pathname is complete in that no other
information is required in order to locate the file that it denotes. A
relative pathname, in contrast, must be interpreted in terms of
information taken from some other pathname. By default the classes in
the java.io package always resolve relative pathnames against the
current user directory. This directory is named by the system property
user.dir, and is typically the directory in which the Java virtual
machine was invoked.
I assumed you were using Eclipse, by default Eclipse sets the user.dir to your the root of your project. From reading other material, Netbeans follows the same convention.
This can be tested with the following code, which should output the path to your project:
System.out.println(System.getProperty("user.dir"));
Placing the file in the root of your directory allows it to be found by the FileInputStream.
If you're using NetBeans (perhaps similar for Eclipse), make sure that your file is in NetbeansProjects/YourProject/
If you have compiled your program to .jar file, put the txt to same place where .jar is.
Either you put the file in the "root directory" of your project or provide the "absolute path" of the file as argument to FileInputStream.
Hope that helps. :).
This way, the program expects the file in the working folder,. i.e. where you execute java. For loading from classpath (Default package), try getResourceAsStream.
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);
So here is my program, which works ok:
import java.io.FileReader;
import java.io.BufferedReader;
import java.io.IOException;
import java.util.Scanner;
import java.util.Locale;
public class ScanSum {
public static void main(String[] args) throws IOException {
Scanner s = null;
double sum = 0;
try {
s = new Scanner(new BufferedReader(new FileReader("D:/java-projects/HelloWorld/bin/usnumbers.txt")));
s.useLocale(Locale.US);
while (s.hasNext()) {
if (s.hasNextDouble()) {
sum += s.nextDouble();
} else {
s.next();
}
}
} finally {
s.close();
}
System.out.println(sum);
}
}
As you can see, I am using absolute path to the file I am reading from:
s = new Scanner(new BufferedReader(new FileReader("D:/java-projects/HelloWorld/bin/usnumbers.txt")));
The problem arises when I try to use the relative path:
s = new Scanner(new BufferedReader(new FileReader("usnumbers.txt")));
I get an error:
Exception in thread "main" java.lang.NullPointerException
at ScanSum.main(ScanSum.java:24)
The file usnumbers.txt is in the same directory as the ScanSum.class file:
D:/java-projects/HelloWorld/bin/ScanSum.class
D:/java-projects/HelloWorld/bin/usnumbers.txt
How could I solve this?
If aioobe#'s suggestion doesn't work for you, and you need to find out which directory the app is running from, try logging the following:
new File(".").getAbsolutePath()
From which directory is the class file executed? (That would be the current working directory and base directory for relative paths.)
If you simply launch the application from eclipse, the project directory will be the working directory, and you should in that case use "bin/usnumbers.txt".
The NullPointerException is due to the fact that new FileReader() expression is throwing a FileNotFoundException, and the variable s is never re-assigned a non-null value.
The file "usnumbers.txt" is not found because relative paths are resolved (as with all programs) relative to the current working directory, not one of the many entries on the classpath.
To fix the first problem, never assign a meaningless null value just to hush the compiler warnings about unassigned variables. Use a pattern like this:
FileReader r = new FileReader(path);
try {
Scanner s = new Scanner(new BufferedReader(r));
...
} finally {
r.close();
}
For the second problem, change directories to the directory that contains "usnumbers.txt" before launching java. Or, move that file to the directory from which java is run.
It must be a FileNotFoundException causing NPE in the finally block.
Eclipse, by default, executes the class with the project folder (D:/java-projects/HelloWorld in your case ) as the working directory. Put the usnumbers.txt file in that folder and try. Or change the working directory in Run Configuration -> Argument tab
Since your working directory is “D:/java-projects/HelloWorld”
#pdbartlett's id is great, But
String filePath = new File(".").getAbsolutePath()
will output "D:/java-projects/HelloWorld/." which is not easy to add your extra relative path like "filePath" + "/src/main/resources/" + FILENAME which located in resources folder.
I suggest with String filePath = new File("").getAbsolutePath() which return the project root folder
In Eclipse, you can also look under "Run Configurations->Than TAB "Classpath".
By default the absolut path is listed under "User Entries" in [icon] 'your.path' (default classpath)
Put the file in resources folder.
ClassLoader classLoader = getClass().getClassLoader();
String file1 = classLoader.getResource("myfile.csv").getPath();