Consider the code sample below. Migrator class takes two input files, processes it and writes the output to final.tbl.
I want final.tbl to be created on the same path where the folder of input files is present.
Also the execute method should take relative path of this generated final.tbl file.
public class Migrator{
public void Migrate(String path1,String path2){
PrintStream out = new PrintStream("final.tbl");//I need relative path as that of input folder path i.e path1,path2
//.....
//.....Processing
}
}
class MainProcess{
public execute(String path){
//here the execute method should the the relative path of above final.tbl file
}
public static void main(String args[]){
}
}
Path path = Paths.get(path1);
PrintStream out = new PrintStream(path.getParent().toString() + "\\final.tbl");
I think you can use getAbsolutePath to get path to your input files:
public class Migrator{
public void Migrate(String path1,String path2){
File f = new File(path1);
String absolutePath = f.getAbsolutePath(); // use absolutePath for your PrintStream
PrintStream out = new PrintStream(absolutePath);//I need relative path as that of input folder path i.e path1,path2
//.....
//.....Processing
}
}
Hope it helped
Use the getParentFile()
File target = new File(new File(path1).getParentFile(), "final.tbl");
PrintStream out = new PrintStream(target);
Related
I have the following prefix:
String prefix = TemplatesReader.class.getClassLoader().getResource("templates/").getPath();
and have method
public byte[] read(String pathToTemplate) {
return Files.readAllBytes(Paths.get(prefix + pathToTemplate));
}
in intellij idea works correctly, but when starting jar an error occurs:
java.nio.file.NoSuchFileException: file:/app.jar!/BOOT-INF/classes!/templates/request-orders/unmarked/RequestOrderUnmarked.pdf
You must not assume that a resource is a file. When the resource is inside a .jar file, it is a part of that .jar file; it is no longer a separate file at all.
You cannot use Files or Paths to read the resource.
You cannot use the getPath() method of URL. It does not return a file name. It only returns the path portion of the URL (that is, everything between the URL’s scheme/authority and its query portion), which is not a file path at all.
Instead, read the resource using getResourceAsStream:
private static final String RESOURCE_PREFIX = "/templates/";
public byte[] read(String pathToTemplate)
throws IOException {
try (InputStream stream = TemplatesReader.class.getResource(
RESOURCE_PREFIX + pathToTemplate)) {
return stream.readAllBytes();
}
}
I have a specified URL that i cant change i.e.
/opt/local/java/config/npvr.properties
where should i place my file so that the following code can work:
String PROPERTIESFILEPATH1 = "/opt/local/java/config/npvr.properties";
File tmPropertiesFile = new File(PROPERTIESFILEPATH);
Properties properties = new Properties();
if (tmPropertiesFile.exists()) {
.....
}
i have tried placing my file in directory shown below but it didn't work:
My problem is that I can change only the location of property-file without changing the code to solve this problem. Please Help.
The file needs to go in /opt/local/java/config, not [projectdir]/opt/local/java/config. You are putting it in the wrong place
If you are using linux then add this file /opt/local/java/config/ directory location outside of your project.
"OR"
in windows C:\opt\local\java\config\ here.
use getClass().getResourceAsStream to load property file from relative of class
public class Main {
public static void main(String args[]) throws Exception{
String PROPERTIESFILEPATH = "/opt/local/java/config/npvr.properties";
//File tmPropertiesFile = new File(PROPERTIESFILEPATH);
Properties properties = new Properties();
InputStream ins=null;
//ins=new FileInputStream(PROPERTIESFILEPATH);
ins=new Main().getClass().getResourceAsStream(PROPERTIESFILEPATH);
properties.load(ins);
System.out.println(properties.get("Hello"));
}
}
I'm trying to get a resource (image.png, in the same package as this code) from a static method using this code:
import java.net.*;
public class StaticResource {
public static void main(String[] args) {
URL u = StaticResource.class.getClass().getResource("image.png");
System.out.println(u);
}
}
The output is just 'null'
I've also tried StaticResource.class.getClass().getClassLoader().getResource("image.png");
, it throws a NullPointerException
I've seen other solutions where this works, what am I doing wrong?
Remove the ".getClass()" part.
Just use
URL u = StaticResource.class.getResource("image.png");
Always try to place the resources outside the JAVA code to make it more manageable and reusable by other package's class.
You can try any one
// Read from same package
URL url = StaticResource.class.getResource("c.png");
// Read from same package
InputStream in = StaticResource.class.getResourceAsStream("c.png");
// Read from absolute path
File file = new File("E:/SOFTWARE/TrainPIS/res/drawable/c.png");
// Read from images folder parallel to src in your project
File file = new File("images/c.jpg");
// Read from src/images folder
URL url = StaticResource.class.getResource("/images/c.png")
// Read from src/images folder
InputStream in = StaticResource.class.getResourceAsStream("/images/c.png")
I'm using apache's FileUpload in order to upload some files to my webserver. Problem is I don't want to upload them to a specific location on the machine i.e : c:\tmp, but rather to a relative path such as /ProjectName/tmp/
Here's my code:
private static final long serialVersionUID = 1L;
private String TMP_DIR_PATH = "c:\\tmp";
private File tmpDir;
private static final String DESTINATION_DIR_PATH ="/files";
private File destinationDir;
public void init(ServletConfig config) throws ServletException {
super.init(config);
tmpDir = new File(TMP_DIR_PATH);
if(!tmpDir.isDirectory()) {
throw new ServletException(TMP_DIR_PATH + " is not a directory");
}
String realPath = getServletContext().getRealPath(DESTINATION_DIR_PATH);
destinationDir = new File(realPath);
if(!destinationDir.isDirectory()) {
throw new ServletException(DESTINATION_DIR_PATH+" is not a directory");
}
}
I want to change TMP_DIR_PATH so it's relative to my project, help would be appreciated!
If your actual concern is the hardcoding of the c:\\tmp part which makes the code unportable, then consider using File#createTempFile() instead. This will create the file in the platform default temp location as specified by the java.io.tmpdir system property.
File file = File.createTempFile("upload", ".tmp");
OutputStream output = new FileOutputStream(file);
// ...
If you want the temp directory to be relative to your project you'll have to use getRealPath and then make sure you create the directory if it doesn't already exist.
You can also use File dir = (File) getServletContext().getAttribute("javax.servlet.context.tempdir"); to fetch an app-specific temp directory provided by the container.
I am trying to create a file using
File newFile = new File("myFile");
However no file called "myFile" is created. This is within a Web application Project i.e. proper form to be pakaged as a WAR but I am calling it as part of a main method (just to see how this works).
How can I make it so that a new file is created at a location relative to the current one i.e not have to put in an absolute path.
EDIT:
newFile.createFile();
Doesn't seem to work:
Here is the entire code:
import java.io.File;
import java.io.IOException;
public class Tester {
public static void main(String[] args) throws IOException{
Tester test = new Tester();
test.makeFile();
}
public void makeFile() throws IOException{
File newFile = new File("myFile");
newFile.createNewFile();
}
}
In answer to your comment. The file will be created in the current directory of the process, unless you specifiy otherwise.
// new file in current directory
File f = new File("yourFile");
f.createNewFile();
System.out.println("Path:" + f.getAbsolutePath());
To create it in a directory of your choosing:
File f = new File("c:\\yourDirectory","yourFile");
f.createNewFile();
System.out.println("Path:" + f.getAbsolutePath());
newFile.createNewFile();
you could use newFile.createNewFile();