Develop Reference

About citing external files in java application

I am writing a java application, in which I am automatically importing external csv files in background to do the computation. But the problem is that I am using "absolute" file path in my java program, the generated jar file will not work in another computer. Is there anyway in java to use a kind of "working directory path" so that I can still run the jar file in another computer as long as I put the csv files I'd like to import in the same folder with the jar file?
Thanks!

You can read a file using its name like
try (BufferedReader br = new BufferedReader(new FileReader("text.txt"))) {
String line;
while ((line=br.readLine())!=null) {
System.out.println(line);
}
} catch (Exception e) {
e.printStackTrace();
}
Here text.txt should be in the same working directory where the jar was executed.
You can also read the directory name from the command line, using the command line arguments like
public static void main(String[] args) {
//check if there were any command line arguments
if (args.length > 0) {
// args[0] is the first command line argument unlike C where args[0] would give u the executable's name
} else {
System.err.println("Usage: java -jar <jar_name> [directory_names..]");
}
}
You can also have a configuration file such as a properties file to read the directory names.

new File(".") give you the relative path
you can write relative path like that :
File file = new File(".\\CSVs\\myfile.csv");

System.getProperty("user.dir") will return you the working directory.
System.getProperty("user.dir")+"\\myfile.txt"
More informations here :system properties, oracle docs

How to create a working .jar for a java project w/ excel input and System.out.print output?

I created this java project that basically gets data from an user determined excel file and uses Syste.out.println() to display the results. It works as I want it to in eclipse, however when I exported is as a .jar file, it doesn't work properly. It prompts for the excel file location to be entered, however, does not display the output, not even an error. I do not know how to do it from the terminal so I'm running it by double-clcking it. Also, I want the user to choose any excel file they want, so what should they write down as the location when prompted to do so? Right now, the excel file is in the same directory as the project. So just the name of the excel file is enough input, but what if it is not in the directory, how do i show it's location then?
Thank you

First of all, in your JAR file the file META-INF/MANIFEST.MF must contain your main class (i.e. the class with the main() method), with a line like this:
Main-Class: mypackage.MyMainClass
Make sure your settings in Eclipse are generating this line when generating the JAR file.
You can run it from the terminal with java -jar yourapp.jar however I presume that you have some extra libraries you need to include in the classpath with the -cp switch.
The working directory is normally the directory from where you are running the application. If you want to specify a different path it has to be either relative to that, or absolute.
An absolute path in windows would be something like: "C:\mydatafolder\myexcelsheet.xls"
An absolute path in Unix would be something like: "/home/myaccount/mydatafolder/myexcelsheet.xls"

You need to read System.in to get the file path, with a function like this:
private static String getFilePath () {
String filePath = null;
while (true) {
System.out.println("Please input the path of the file:");
try {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
filePath = br.readLine();
File file = new File(filePath);
if (!file.exists()) {
System.out.println("Sorry, invalid file path. Please try again.");
} else {
return filePath;
}
} catch (IOException e) {
e.printStackTrace();
System.out.println("Sorry, unexpected error happened, Please try again.");
}
}
}

java.io.FileNotFoundException (File not found) using Scanner. What's wrong in my code?

I've a .txt file ("file.txt") in my netbeans "/build/classes" directory.
In the same directory there is the .class file compiled for the following code:
try {
File f = new File("file.txt");
Scanner sc = new Scanner(f);
}
catch (IOException e) {
System.out.println(e);
}
Debugging the code (breakpoint in "Scanner sc ..") an exception is launched and the following is printed:
java.io.FileNotFoundException: file.txt (the system can't find the
specified file)
I also tried using "/file.txt" and "//file.txt" but same result.
Thank you in advance for any hint

If you just use new File("pathtofile") that path is relative to your current working directory, which is not at all necessarily where your class files are.
If you are sure that the file is somewhere on your classpath, you could use the following pattern instead:
URL path = ClassLoader.getSystemResource("file.txt");
if(path==null) {
//The file was not found, insert error handling here
}
File f = new File(path.toURI());

The JVM will look for the file in the current working directory.
Where this is depends on your IDE settings (how your program is executed).
To figure out where it expects file.txt to be located, you could do
System.out.println(new File("."));
If it for instance outputs
/some/path/project/build
you should place file.txt in the build directory (or specify the proper path relative to the build directory).

Try:
File f = new File("./build/classes/file.txt");

Use "." to denote the current directory
String path = "./build/classes/file.txt";
File f = new File(path);

File Object loads, looking for match in its current directory.... which is Directly in Your project folder where your class files are loaded not in your source ..... put the file directly in the project folder

How to read file from relative path in Java project? java.io.File cannot find the path specified

I have a project with 2 packages:
tkorg.idrs.core.searchengines
tkorg.idrs.core.searchengines
In package (2) I have a text file ListStopWords.txt, in package (1) I have a class FileLoadder. Here is code in FileLoader:
File file = new File("properties\\files\\ListStopWords.txt");
But I have this error:
The system cannot find the path specified
Can you give a solution to fix it?

If it's already in the classpath, then just obtain it from the classpath instead of from the disk file system. Don't fiddle with relative paths in java.io.File. They are dependent on the current working directory over which you have totally no control from inside the Java code.
Assuming that ListStopWords.txt is in the same package as your FileLoader class, then do:
URL url = getClass().getResource("ListStopWords.txt");
File file = new File(url.getPath());
Or if all you're ultimately after is actually an InputStream of it:
InputStream input = getClass().getResourceAsStream("ListStopWords.txt");
This is certainly preferred over creating a new File() because the url may not necessarily represent a disk file system path, but it could also represent virtual file system path (which may happen when the JAR is expanded into memory instead of into a temp folder on disk file system) or even a network path which are both not per definition digestable by File constructor.
If the file is -as the package name hints- is actually a fullworthy properties file (containing key=value lines) with just the "wrong" extension, then you could feed the InputStream immediately to the load() method.
Properties properties = new Properties();
properties.load(getClass().getResourceAsStream("ListStopWords.txt"));
Note: when you're trying to access it from inside static context, then use FileLoader.class (or whatever YourClass.class) instead of getClass() in above examples.

The relative path works in Java using the . specifier.
. means same folder as the currently running context.
.. means the parent folder of the currently running context.
So the question is how do you know the path where the Java is currently looking?
Do a small experiment
File directory = new File("./");
System.out.println(directory.getAbsolutePath());
Observe the output, you will come to know the current directory where Java is looking. From there, simply use the ./ specifier to locate your file.
For example if the output is
G:\JAVA8Ws\MyProject\content.
and your file is present in the folder "MyProject" simply use
File resourceFile = new File("../myFile.txt");
Hope this helps.

The following line can be used if we want to specify the relative path of the file.
File file = new File("./properties/files/ListStopWords.txt");

InputStream in = FileLoader.class.getResourceAsStream("<relative path from this class to the file to be read>");
try {
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String line = null;
while ((line = reader.readLine()) != null) {
System.out.println(line);
}
} catch (Exception e) {
e.printStackTrace();
}

try .\properties\files\ListStopWords.txt

I could have commented but I have less rep for that.
Samrat's answer did the job for me. It's better to see the current directory path through the following code.
File directory = new File("./");
System.out.println(directory.getAbsolutePath());
I simply used it to rectify an issue I was facing in my project. Be sure to use ./ to back to the parent directory of the current directory.
./test/conf/appProperties/keystore

While the answer provided by BalusC works for this case, it will break when the file path contains spaces because in a URL, these are being converted to %20 which is not a valid file name. If you construct the File object using a URI rather than a String, whitespaces will be handled correctly:
URL url = getClass().getResource("ListStopWords.txt");
File file = new File(url.toURI());

Assuming you want to read from resources directory in FileSystem class.
String file = "dummy.txt";
var path = Paths.get("src/com/company/fs/resources/", file);
System.out.println(path);
System.out.println(Files.readString(path));
Note: Leading . is not needed.

I wanted to parse 'command.json' inside src/main//js/Simulator.java. For that I copied json file in src folder and gave the absolute path like this :
Object obj = parser.parse(new FileReader("./src/command.json"));

For me actually the problem is the File object's class path is from <project folder path> or ./src, so use File file = new File("./src/xxx.txt"); solved my problem

For me it worked with -
String token = "";
File fileName = new File("filename.txt").getAbsoluteFile();
Scanner inFile = null;
try {
inFile = new Scanner(fileName);
} catch (FileNotFoundException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
while( inFile.hasNext() )
{
String temp = inFile.next( );
token = token + temp;
}
inFile.close();
System.out.println("file contents" +token);

If text file is not being read, try using a more closer absolute path (if you wish
you could use complete absolute path,) like this:
FileInputStream fin=new FileInputStream("\\Dash\\src\\RS\\Test.txt");
assume that the absolute path is:
C:\\Folder1\\Folder2\\Dash\\src\\RS\\Test.txt

String basePath = new File("myFile.txt").getAbsolutePath();
this basepath you can use as the correct path of your file

if you want to load property file from resources folder which is available inside src folder, use this
String resourceFile = "resources/db.properties";
InputStream resourceStream = ClassLoader.getSystemClassLoader().getResourceAsStream(resourceFile);
Properties p=new Properties();
p.load(resourceStream);
System.out.println(p.getProperty("db"));
db.properties files contains key and value db=sybase

If you are trying to call getClass() from Static method or static block, the you can do the following way.
You can call getClass() on the Properties object you are loading into.
public static Properties pathProperties = null;
static {
pathProperties = new Properties();
String pathPropertiesFile = "/file.xml";
// Now go for getClass() method
InputStream paths = pathProperties.getClass().getResourceAsStream(pathPropertiesFile);
}

How to pass a text file as a argument?

Im trying to write a program to read a text file through args but when i run it, it always says the file can't be found even though i placed it inside the same folder as the main.java that im running.
Does anyone know the solution to my problem or a better way of reading a text file?

Do not use relative paths in java.io.File.
It will become relative to the current working directory which is dependent on the way how you run the application which in turn is not controllable from inside your application. It will only lead to portability trouble. If you run it from inside Eclipse, the path will be relative to /path/to/eclipse/workspace/projectname. If you run it from inside command console, it will be relative to currently opened folder (even though when you run the code by absolute path!). If you run it by doubleclicking the JAR, it will be relative to the root folder of the JAR. If you run it in a webserver, it will be relative to the /path/to/webserver/binaries. Etcetera.
Always use absolute paths in java.io.File, no excuses.
For best portability and less headache with absolute paths, just place the file in a path covered by the runtime classpath (or add its path to the runtime classpath). This way you can get the file by Class#getResource() or its content by Class#getResourceAsStream(). If it's in the same folder (package) as your current class, then it's already in the classpath. To access it, just do:
public MyClass() {
URL url = getClass().getResource("filename.txt");
File file = new File(url.getPath());
InputStream input = new FileInputStream(file);
// ...
}
or
public MyClass() {
InputStream input = getClass().getResourceAsStream("filename.txt");
// ...
}

Try giving an absolute path to the filename.
Also, post the code so that we can see what exactly you're trying.

When you are opening a file with a relative file name in Java (and in general) it opens it relative to the working directory.
you can find the current working directory of your process using
String workindDir = new File(".").getAbsoultePath()
Make sure you are running your program from the correct directory (or change the file name so that it will be relative to where you are running it from).

If you're using Eclipse (or a similar IDE), the problem arises from the fact that your program is run from a few directories above where the actual source is located. Try moving your file up a level or two in the project tree.
Check out this question for more detail.

The simplest solution is to create a new file, then see where the output file is. That is the correct place to put your input file into.

If you put the file and the class working with it under same package can you use this:
Class A {
void readFile (String fileName) {
Url tmp = A.class.getResource (fileName);
// Or Url tmp = this.getClass().getResource (fileName);
File tmpFile = File (tmp);
if (tmpFile.exists())
System.out.print("I found the file.")
}
}
It will help if you read about classloaders.

say I have a text file input.txt which is located on the desktop
and input.txt has the following content
i came
i saw
i left
and below is the java code for reading that text file
public class ReadInputFromTextFile {
public static void main(String[] args) throws Exception
{
File file = new File(
"/Users/viveksingh/desktop/input.txt");
BufferedReader br
= new BufferedReader(new FileReader(file));
String st;
while ((st = br.readLine()) != null)
System.out.println(st);
}
}
output on the console:
i came
i saw
i left