This code throws an exception for the line Integer.parseInt(data[2])

Question

In my playerList.txt I have lines constructed as such:

Nathaniel, Clyne, 12, first

Why can't it accept Integer.parseInt?

I have a class MemberPlayer that defines a MemberPlayer should have the first name, last name, age and team (first or second).

Exception in thread "main" java.lang.NumberFormatException: For input string: " 12"
at java.base/java.lang.NumberFormatException.forInputString NumberFormatException.java:65)
at java.base/java.lang.Integer.parseInt(Integer.java:638)
at java.base/java.lang.Integer.parseInt(Integer.java:770)
at FileHandler.getContent(FileHandler.java:53)
at Main.main(Main.java:55)

   public static final String FINAL_PATH = "playerList.txt";

   public static ArrayList<MemberPlayer> getContent(String FINAL_PATH)
     {
        ArrayList<MemberPlayer> list = new ArrayList<MemberPlayer>();
        try
        {
              Scanner sc = new Scanner(new File(FINAL_PATH));
              while(sc.hasNextLine())
              {                                   
                    String data[] = sc.nextLine().split(",");
                    list.add(new MemberPlayer(data[0], data[1], 
                    Integer.parseInt(data[2]), data[3]));
              }

        }
        catch(FileNotFoundException e)
        {

        }
        return list;
   }

looks like you need to trim the leading blank from the string value `" 12"`. — Filburt, Sep 24 '18 at 11:37
[Converting a String with spaces to an Integer in java](https://stackoverflow.com/questions/23851668/converting-a-string-with-spaces-to-an-integer-in-java) (although I'd say you should be fixing your regex instead of doing this) — Bernhard Barker, Sep 24 '18 at 12:33

Andrew Tobilko · Accepted Answer · 2018-09-24T11:44:52.643

2

" 12" can't be parsed into a valid int because it has a leading space, but "12" can:

Integer.parseInt(data[2].trim())

Make sure that all of data[0] ... data[3] exist before accessing them. The file might be corrupted or inaccurately filled.

edited Sep 24 '18 at 11:44

answered Sep 24 '18 at 11:37

Andrew Tobilko

48,120
14
91
142

score 1 · Answer 2 · answered Sep 24 '18 at 11:40

1

Try to split with ", " or ",\\s".

answered Sep 24 '18 at 11:40

dnsiv

520
4
20

1

Better even `.split(",\\s*", 4)`. – Joop Eggen Sep 24 '18 at 11:44
Better yet - split on comma and trim before you use. – duffymo Sep 24 '18 at 11:51

This code throws an exception for the line Integer.parseInt(data[2])

2 Answers2