如何使用parseInt将字符串转换为整数？

Question

I have this code:

public static void main(String[] args) {
        List<Valtozas> lista = new ArrayList<Valtozas>();

        try {
            File fajl = new File("c://data//uzemanyag.txt");
            Scanner szkenner = new Scanner(fajl, "UTF8");

            while (szkenner.hasNext()) {
                String sor = szkenner.nextLine();
                String [] darabok = sor.split(";");
                String szoveg = darabok[0];
                Valtozas valtozas = new Valtozas(Integer.valueOf(darabok[0]), Integer.valueOf(darabok[1]), Integer.valueOf(darabok[2]));
                lista.add(valtozas);
            }
        }
        catch (Exception e) {

        }


        //3.FELADAT
        System.out.println("3.Feladat: Változások száma: "+lista.size());
    }

}

Here I want to convert the String to int, but I cant. I tried the Integer.Valueof(darabok[0]), but its not working. And nothing is happening, so the code is build, but quit from the while.

Answer 1

Based on the source code you have shown us, the problem is that there is a format mismatch between the input file and the program you are trying to read.

The program reads the file a line at a time, and splits it into fields using a single semicolon (with no whitespace!) as the file separator. Then it tries to parse the first three fields of each split line as integers.

For this to work the following must be true:

• Every line must have at least three fields. (Otherwise you will get a ArrayIndexOutOfBound exception.)
• The three fields must match the following regex: -?[0-9]+ ie an optional minus signed followed by one or more decimal digits.
• The resulting number must be between Integer.MIN_VALUE and Integer.MAX_VALUE .

Elaborating on the above:

• Leading and trailing whitespace characters are not allowed.
• Embedded decimals markers and grouping characters are not allowed.
• Scientific notation is not allowed.
• Numbers that are too large or too small are not allowed.

Note that if any of the above constraints is not met, then the runtime system will throw an exception. Unfortunately, you surround your code with this:

    try {
        ...
    }
    catch (Exception e) {

    }

That basically ignores all exceptions by catching them and doing nothing in the handler block. You are saying to Java "if anything goes wrong, don't tell me about it!". So, naturally, Java doesn't tell you what is going wrong.

DON'T EVER DO THIS . This is called exception squashing, and it is a really bad idea ¹ .

There are two ways to address this:

1. Print or log the stacktrace; eg

     catch (Exception e) { e.printStackTrace(); } 

2. Remove the try / catch, and add throws IOException to the signature of your main method.

(You can't just remove the try / catch because new Scanner(fajl, "UTF8") throws IOException . That's a checked exception so must be handled or declared in the method signature.)

Once you have dealt with the exception properly you will get an exception message and stacktrace that tells you what is actually going wrong. Read it, understand it, and fix your program.

---

^{1 - It is like taping over the "annoying" red light that indicates that your car's oil level is low!}