如何使Java从文本中读取数字列表？

Question

How do I get my Java code to read the next number from a list of numbers in a text file. My output repeats the first number multiple times, how do I fix this?

public static void main(String[] args) throws Exception {
    for (int l = 0; l < 9; l++) {
        java.io.File myfile;
        String mypath;
        mypath = "/Users/tonyg/Downloads";
        myfile = new java.io.File(mypath + "/file.txt");
        Scanner myinfile = new Scanner(myfile);
        int val1;
        val1 = myinfile.nextInt();
        System.out.println(val1);
    }
}

OUTPUT:

385

385

385

385

385

385

385

385

385

Answer 1

It looks like your initializing the Scanner on every iterations of the for loop by simply initializing before the loop, your issue should be resolved. It's also best practice to close the Scanner after using the resource

public static void main(String[] args) throws Exception {
    java.io.File myfile;
    String mypath;
    mypath = "/Users/tonyg/Downloads";
    myfile = new java.io.File(mypath + "/file.txt");
    Scanner myinfile = new Scanner(myfile);
    for (int l = 0; l < 9; l++) {
        int val1;
        val1 = myinfile.nextInt();
        System.out.println(val1);
    }

Answer 2

1. Initialize Scanner out of loop
2. Catch FileNotFoundException
3. Combine declaration and initialization of variables (in this particular case)
4. Use clear camelCase identifiers for your variables
5. Avoid using l (lowercase L ) as a variable identifier. In many fonts l (lowercase L ) and 1 (digit one) look similar. This may lead to future bugs caused by typos.
6. Close your Scanner in the end

(#1, #2 and #6 can be achieved by using try-with-resources )

String dirPath = "/Users/tonyg/Downloads";
String filePath = dirPath + "/file.txt";
int count = 9;

try(Scanner scanner = new Scanner(new File(filePath))){
    for (int i = 0; i < count; i++) {
        int value = scanner.nextInt();
        System.out.println(value);
    }
} catch (FileNotFoundException e){
    // Print stack-trace or do something else
    e.printStackTrace();
}