I have a desktop application using Swing library. Application is running a batch file. So I created a lib folder in main project directory and put batch file in it. To run this, I am showing lib\a.exe to run this. It is working on my laptop. I exported .jar and put lib folder next to it. It is working on my laptop, but not working on some other laptops. How to fix this?
Error message is: Windows cannot found lib\a.exe.
String command = "cmd /c start lib\\a.exe";
try {
Runtime.getRuntime().exec(command);
increaseProgressBarValue();
} catch (IOException e) {
e.printStackTrace();
}
You need two things:
find the directory of the jar file of your application
call a.exe with the correct working directory
You can get the location of the jar with the getJar method below:
private static File getJar(Class clazz) throws MalformedURLException {
String name = clazz.getName().replace('.','/') + ".class";
ClassLoader cl = Thread.currentThread().getContextClassLoader();
URL url = cl.getResource(name);
System.out.println(url);
if (!"jar".equals(url.getProtocol())) {
throw new MalformedURLException("Expected a jar: URL " + url);
}
String file = url.getPath();
int pos = file.lastIndexOf('!');
if (pos < 0) {
throw new MalformedURLException("Expected ! " + file);
}
url = new URL(file.substring(0, pos));
if (!"file".equals(url.getProtocol())) {
throw new MalformedURLException("Expected a file: URL " + url);
}
String path = url.getPath();
if (path.matches("/[A-Za-z]:/")) { // Windoze drive letter
path = path.substring(1);
}
return new File(path);
}
To call lib\a.exe, you can do something like this:
File jar = getJar(MyClass.class); // MyClass can be any class in you jar file
File dir = jar.getParentFile();
ProcessBuilder builder = new ProcessBuilder();
builder.command("lib\\a.exe");
builder.directory(dir);
...
Process p = builder.start();
...
Maybe you have to try if this folder lib exists and if it doesn't than create it with
file.mkdir();
This is a just a checking. But your filepath must be like this ../lib/a.exe.
Related
I have a spring-boot application and I have some files placed inside /src/main/java/resources which I am trying to read it in my code. When the same code tries to read from docker container it says file does not exist. The same code works perfectly fine via localhost
The files are under /src/main/java/resources/data folder and this is my code which tries to read the file
private String getJson(String folderName, String id, StringBuilder sb) throws Exception {
String responseJson = null;
String filePath = "data" + File.separator + folderName + File.separator + id + ".json";
LOG.info("printing filePath : " + filePath);
LOG.info("printing id : " + id);
File f = new File(filePath);
// if(f.exists()){
try (InputStream inputStream = this.getClass().getClassLoader().getResourceAsStream(filePath)) {
LOG.info("printing inputStream : " + inputStream);
if (inputStream != null) {
responseJson = IOUtils.toString(inputStream, StandardCharsets.UTF_8.name());
}
if (responseJson == null || responseJson.isEmpty()) {
LOG.info("json response is null : ");
throw new JsonNotFoundException(Constant.JSON_NOT_FOUND);
}
sb.append(responseJson);
} catch (IOException e) {
LOG.info("IO exception : ");
throw new IOException(e);
} catch (Exception e) {
LOG.info(" exception : ");
throw new Exception(e);
}
// }
// else{
// LOG.info("file doesnt exists : " + filePath);
// }
return sb.toString();
}
An example for the file path : src/main/resources/data/get-products/1420-17612-82.json
Docker file content
{
"commands":
[
"rm -rf .tmp",
"git clone git#github.com:{orgnname}/{test-service.git} -b COECP-973-Configure-logging-mechanism-for-Service .tmp/test-service",
"docker build .tmp/test-service/.docker/build/db -t local/test-service/db",
"docker build .tmp/test-service -t local/test-service/app"
]
}
So... you messed path for File and for resources from class path. What is the reason to have File f = new File(filePath);?
Here are things:
If you use File - files must be available to the JVM and as long as you use relative path like data\folderxxx\filexxx.json it must be available in container file system. I.e. data folder must be placed in image or mounted from outside exactly into directory from where JVM runs
If you use ClassLoader and ResourceAsStream root of your data directory must be defined in class path for JVM or be in Jar file - it is a root in classpath as well. Check your jar file - if data directory is in root of jar - all fine and files will be available by this.getClass().getClassLoader().getResourceAsStream(filePath), but not for new File(filePath)!
if not - make it happen or update your filePath for ResourceAsStream accordingly.
I have had faced the same issue earlier, though our requirement got more complex over time, but the following code should solve your problem:
ClassPathResource cp = new ClassPathResource("relative_path_to_file");
File f = null;
if (cp.exists())
f = cp.getFile();
The condition is if the directory exists it has to create files in that specific directory without creating a new directory.
The below code only creates a file with the new directory but not for the existing directory . For example the directory name would be like "GETDIRECTION":
String PATH = "/remote/dir/server/";
String fileName = PATH.append(id).concat(getTimeStamp()).append(".txt");
String directoryName = PATH.append(this.getClassName());
File file = new File(String.valueOf(fileName));
File directory = new File(String.valueOf(directoryName));
if (!directory.exists()) {
directory.mkdir();
if (!file.exists() && !checkEnoughDiskSpace()) {
file.getParentFile().mkdir();
file.createNewFile();
}
}
FileWriter fw = new FileWriter(file.getAbsoluteFile());
BufferedWriter bw = new BufferedWriter(fw);
bw.write(value);
bw.close();
Java 8+ version:
Files.createDirectories(Paths.get("/Your/Path/Here"));
The Files.createDirectories() creates a new directory and parent directories that do not exist. This method does not throw an exception if the directory already exists.
This code checks for the existence of the directory first and creates it if not, and creates the file afterwards. Please note that I couldn't verify some of your method calls as I don't have your complete code, so I'm assuming the calls to things like getTimeStamp() and getClassName() will work. You should also do something with the possible IOException that can be thrown when using any of the java.io.* classes - either your function that writes the files should throw this exception (and it be handled elsewhere), or you should do it in the method directly. Also, I assumed that id is of type String - I don't know as your code doesn't explicitly define it. If it is something else like an int, you should probably cast it to a String before using it in the fileName as I have done here.
Also, I replaced your append calls with concat or + as I saw appropriate.
public void writeFile(String value){
String PATH = "/remote/dir/server/";
String directoryName = PATH.concat(this.getClassName());
String fileName = id + getTimeStamp() + ".txt";
File directory = new File(directoryName);
if (! directory.exists()){
directory.mkdir();
// If you require it to make the entire directory path including parents,
// use directory.mkdirs(); here instead.
}
File file = new File(directoryName + "/" + fileName);
try{
FileWriter fw = new FileWriter(file.getAbsoluteFile());
BufferedWriter bw = new BufferedWriter(fw);
bw.write(value);
bw.close();
}
catch (IOException e){
e.printStackTrace();
System.exit(-1);
}
}
You should probably not use bare path names like this if you want to run the code on Microsoft Windows - I'm not sure what it will do with the / in the filenames. For full portability, you should probably use something like File.separator to construct your paths.
Edit: According to a comment by JosefScript below, it's not necessary to test for directory existence. The directory.mkdir() call will return true if it created a directory, and false if it didn't, including the case when the directory already existed.
Trying to make this as short and simple as possible. Creates directory if it doesn't exist, and then returns the desired file:
/** Creates parent directories if necessary. Then returns file */
private static File fileWithDirectoryAssurance(String directory, String filename) {
File dir = new File(directory);
if (!dir.exists()) dir.mkdirs();
return new File(directory + "/" + filename);
}
I would suggest the following for Java8+.
/**
* Creates a File if the file does not exist, or returns a
* reference to the File if it already exists.
*/
public File createOrRetrieve(final String target) throws IOException {
final File answer;
Path path = Paths.get(target);
Path parent = path.getParent();
if(parent != null && Files.notExists(parent)) {
Files.createDirectories(path);
}
if(Files.notExists(path)) {
LOG.info("Target file \"" + target + "\" will be created.");
answer = Files.createFile(path).toFile();
} else {
LOG.info("Target file \"" + target + "\" will be retrieved.");
answer = path.toFile();
}
return answer;
}
Edit: Updated to fix bug as indicated by #Cataclysm and #Marcono1234. Thx guys:)
code:
// Create Directory if not exist then Copy a file.
public static void copyFile_Directory(String origin, String destDir, String destination) throws IOException {
Path FROM = Paths.get(origin);
Path TO = Paths.get(destination);
File directory = new File(String.valueOf(destDir));
if (!directory.exists()) {
directory.mkdir();
}
//overwrite the destination file if it exists, and copy
// the file attributes, including the rwx permissions
CopyOption[] options = new CopyOption[]{
StandardCopyOption.REPLACE_EXISTING,
StandardCopyOption.COPY_ATTRIBUTES
};
Files.copy(FROM, TO, options);
}
Simple Solution using using java.nio.Path
public static Path createFileWithDir(String directory, String filename) {
File dir = new File(directory);
if (!dir.exists()) dir.mkdirs();
return Paths.get(directory + File.separatorChar + filename);
}
If you create a web based application, the better solution is to check the directory exists or not then create the file if not exist. If exists, recreate again.
private File createFile(String path, String fileName) throws IOException {
ClassLoader classLoader = getClass().getClassLoader();
File file = new File(classLoader.getResource(".").getFile() + path + fileName);
// Lets create the directory
try {
file.getParentFile().mkdir();
} catch (Exception err){
System.out.println("ERROR (Directory Create)" + err.getMessage());
}
// Lets create the file if we have credential
try {
file.createNewFile();
} catch (Exception err){
System.out.println("ERROR (File Create)" + err.getMessage());
}
return file;
}
A simple solution using Java 8
public void init(String multipartLocation) throws IOException {
File storageDirectory = new File(multipartLocation);
if (!storageDirectory.exists()) {
if (!storageDirectory.mkdir()) {
throw new IOException("Error creating directory.");
}
}
}
If you're using Java 8 or above, then Files.createDirectories() method works the best.
I'm trying to access a resource from a jar file. The resource is located in the same directory where is the jar.
my-dir:
tester.jar
test.jpg
I tried different things including the following, but every time the input stream is null:
[1]
String path = new File(".").getAbsolutePath();
InputStream inputStream = this.getClass().getResourceAsStream(path.replace("\\.", "\\") + "test.jpg");
[2]
File f = new File(this.getClass().getProtectionDomain().getCodeSource().getLocation().toURI().getPath());
InputStream inputStream = this.getClass().getResourceAsStream(f.getParent() + "test.jpg");
Can you give me some hints? Thanks.
If you are sure, that your application's current folder is the folder of the jar, you can simply call InputStream f = new FileInputStream("test.jpg");
The getResource methods will load stuff using the classloader, not through filesystem. This is why your approach (1) failed.
If the folder containing your *.jar and image file is in the classpath, you can get the image resource as if it was on the default-package:
class.getClass().getResourceAsStream("/test.jpg");
Beware: The image is now loaded in the classloader, and as long as the application runs, the image is not unloaded and served from memory if you load it again.
If the path containing the jar file is not given in the classpath, your approach to get the jarfile path is good.
But then simply access the file directly through the URI, by opening a stream on it:
URL u = this.getClass().getProtectionDomain().getCodeSource().getLocation();
// u2 is the url derived from the codesource location
InputStream s = u2.openStream();
Use this tutorial to help you create a URL to a single file in a jar file.
Here's an example:
String jarPath = "/home/user/myJar.jar";
String urlStr = "jar:file://" + jarPath + "!/test.jpg";
InputStream is = null;
try {
URL url = new URL(urlStr);
is = url.openStream();
Image image = ImageIO.read(is);
}
catch(Exception e) {
e.printStackTrace();
}
finally {
try {
is.close();
} catch(Exception IGNORE) {}
}
I have a spring boot web application which I run using java -jar application.jar. I need to get the jar parent folder path dynamically from the code. How can I accomplish that?
I have already tried this, but without success.
Well, what have worked for me was an adaptation of this answer.
The code is:
if you run using java -jar myapp.jar dirtyPath will be something close
to this: jar:file:/D:/arquivos/repositorio/myapp/trunk/target/myapp-1.0.3-RELEASE.jar!/BOOT-INF/classes!/br/com/cancastilho/service.
Or if you run from Spring Tools Suit, something like this:
file:/D:/arquivos/repositorio/myapp/trunk/target/classes/br/com/cancastilho/service
public String getParentDirectoryFromJar() {
String dirtyPath = getClass().getResource("").toString();
String jarPath = dirtyPath.replaceAll("^.*file:/", ""); //removes file:/ and everything before it
jarPath = jarPath.replaceAll("jar!.*", "jar"); //removes everything after .jar, if .jar exists in dirtyPath
jarPath = jarPath.replaceAll("%20", " "); //necessary if path has spaces within
if (!jarPath.endsWith(".jar")) { // this is needed if you plan to run the app using Spring Tools Suit play button.
jarPath = jarPath.replaceAll("/classes/.*", "/classes/");
}
String directoryPath = Paths.get(jarPath).getParent().toString(); //Paths - from java 8
return directoryPath;
}
EDIT:
Actually, if your using spring boot, you could just use the ApplicationHome class like this:
ApplicationHome home = new ApplicationHome(MyMainSpringBootApplication.class);
home.getDir(); // returns the folder where the jar is. This is what I wanted.
home.getSource(); // returns the jar absolute path.
File file = new File(".");
logger.debug(file.getAbsolutePath());
This worked for me to get the path where my jar is running, I hope this is what you are expecting.
Try this code
public static String getParentRealPath(URI uri) throws URISyntaxException {
if (!"jar".equals(uri.getScheme()))
return new File(uri).getParent();
do {
uri = new URI(uri.getSchemeSpecificPart());
} while ("jar".equals(uri.getScheme()));
File file = new File(uri);
do {
while (!file.getName().endsWith(".jar!"))
file = file.getParentFile();
String path = file.toURI().toString();
uri = new URI(path.substring(0, path.length() - 1));
file = new File(uri);
} while (!file.exists());
return file.getParent();
}
URI uri = clazz.getProtectionDomain().getCodeSource().getLocation().toURI();
System.out.println(getParentRealPath(uri));
I am trying to run an executable jar places in the resources folder of my project. If I place the jar in any directory of my File System and provide the absolute path, it works fine.
Please see the code below:
String jarPath = "C:\\JarFolder\\myJar.jar";
String command = "java -jar";
String space = " ";
String params = "-a abc";
try {
proc = Runtime
.getRuntime()
.exec(command + space + jarPath + space + params);
} catch (IOException e) {
e.printStackTrace();
}
But when I place the jar inside the resources folder, and set the relative jar path as:
String jarPath = "..\\..\\..\\resources\\myJar.jar";
I get an error: Error: Unable to access jarfile ..\\..\\..\\resources\\myJar.jar
I have verified the path, it is valid.
Am I doing something wrong here? Is this the correct way to do this?
Use the ClassLoader to get the path of your resource.
String jarPath = this.getClass().getClassLoader().getResource("myJar.jar").getPath();
String command = "java -jar";
String space = " ";
String params = "-a abc";
try {
proc = Runtime
.getRuntime()
.exec(command + space + jarPath + space + params);
} catch (IOException e) {
e.printStackTrace();
}
If this is being ran from a main static method, then just replace this.getClass() with YourClass.class.
Relative path should work straight forward. Need more details of the error description you see and project folder structure (where the executing jar is placed and where is the import folder placed.)
Other alternate solution is to find the absolute file location of the currently executing jar file. You can fetch it with below code snippet.
MyClass.class.getProtectionDomain().getCodeSource().getLocation().toURI().getPath();
Other jar file to be executed must be placed somewhere in the same folder or in some child folders.
Now remove the current jar file name from the absolute path and append the relative path of the other jar file and execute it. It should work.
If you are on window, you can use as follow. I have tested this code and it works.
public class test {
public static void main(String[]args) throws IOException{
String jarPath = test.class.getClass().getResource("/resources/b.jar").getPath();
System.out.println("jarPath "+ jarPath);
//the result path have extra "/" so we have to remove it as follow.
jarPath = jarPath.substring(1);
//and again the result is encoded so we need to decode it back
jarPath = URLDecoder.decode(jarPath);
Runtime
.getRuntime()
.exec("java -jar \""+jarPath+"\"");
}
}
Note: I put my runnable jar in my resources folder with name "b.jar".
And the above codes need some modification to meet your needs.
Good Luck