如何使用命令行 arguments 作为 Java 中的 Scanner 参数

Question

I am writing a program that requires that I use java command line arguments to open a file and find some information within that file. The files we want to access already exist in my projects file directory. The way I am trying to access them is by using the Scanner and just passing args[0] which should be the name of the file that we are trying to access.

Below is the code that I have come up with so far. Right now my code will always throw

java.io.FileNotFoundException: {The Files Name} (The system cannot find the file specified)

public static void main(String[] args) {
    try {
        File inputedFile = new File(args[0]);
        Scanner fileReader = new Scanner(inputedFile);
        while (fileReader.hasNextLine()) {
            String fileData = fileReader.nextLine();
            System.out.println(fileData);
        }
        fileReader.close();
    } catch (FileNotFoundException e) {
        System.out.println("An error occurred when trying to open the file.");
        e.printStackTrace();
    }
}

I believe the reason I am getting this exception is because Java is not able to find the file off of just the file's name. Java needs the location of the file on my system for it to work so for example:

D:\coding-stuff\Java\CS-2420\src\comprehensive\super_simple.txt

would work while

super_simple.txt

would not.

This is an issue for me because I need this code to be run on multiple machines not just my own. So my question is how can I create a "relative link" to the file within my file structure?

Answer 1

Either you know the location of the file or you can obtain the location of the file or you can search for the file.

The System properties contain the path to the working directory, the user's home directory and the temporary directory, so you could place your file in one of those three. Alternatively, you can use the -D option of the java command to define your own [System] property.

You could also define an environment variable and obtain its value via method getenv .

Another option would be to make the file a resource .

If you don't know or can't obtain the path to the file, then you can search for it, either using the Java API or, via class ProcessBuilder , execute an operating system command to search for the file. [Note that the System properties can also tell you the host operating system.]