java.lang.NumberFormatException: For input string: " "

Question

Exception in thread "main" java.lang.NumberFormatException: For input string: "    " 
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)
at java.lang.Long.parseLong(Long.java:430)
at java.lang.Long.valueOf(Long.java:540)
at PckUtiles.lexec.leer(lexec.java:62)
at PckUtiles.lexec.verificar(lexec.java:34)
at PckjForms.Main.main(Main.java:40)

*I have next error when run project "Exception in thread "main" java.lang.NumberFormatException: For input string: " the function of my class is to avoid re-run the application. could help to locate the fault. thank you very much *

here is my class lexec

public class lexec {
    private String ruta = System.getProperties().getProperty("user.dir");
    private File archivo = new File(ruta + "\\Sifme.tmp");
    private int contador = 20;

    public lexec(){};

    public boolean verificar(){
        if(archivo.exists()){
            long time = leer();
            long res = rTiempo(time);
            if(res<contador){
                JOptionPane.showMessageDialog(null, "La aplicación esta en ejecución");
                System.exit(0);
                return false;
            }else{
                tarea_();
                return true;
            }
        }else{
            sifme();
            tarea_();
            return true;
        }
    }

    public long leer(){
        String line = "0";
        BufferedReader br;
        try{
            br = new BufferedReader(new FileReader(archivo));
            while(br.ready()){
                line = br.readLine();
            }
        }catch(IOException e){
            System.err.println(e.getMessage());
        }
        return Long.valueOf(line).longValue();
    }
    public void tarea_(){
        ScheduledExecutorService ses = Executors.newSingleThreadScheduledExecutor();
        ses.scheduleAtFixedRate(
                new Runnable(){
                    @Override
                    public void run(){
                        sifme();
                    }
                },1000,contador*1000,TimeUnit.MILLISECONDS);
    }

    public void sifme(){
        Date fecha = new Date();
        try{
            BufferedWriter bw = new BufferedWriter(new FileWriter(archivo));
            bw.write(String.valueOf(fecha.getTime()));
            bw.close();
        }catch(IOException e){
            System.out.println(e.getMessage());
        }
    }
    public long rTiempo(long tiempoN){
        Date fecha = new Date();
        long t1 = fecha.getTime();
        long tiempo = t1 - tiempoN;
        tiempo = tiempo/1000;
        return tiempo;
    }
    public void detruir_(){
        if(archivo.exists()){
            archivo.delete();
            System.exit(0);
        }
    }
}

@Prateek but previous topic was deleted by the author. So no duplication from now ;-) — Emil Sierżęga
– Emil Sierżęga, Commented Mar 6, 2014 at 11:08

Daniel Kaplan · Accepted Answer · 2013-11-06 00:30:12Z

4

Although you didn't tell us what line gives us the error (even though you should have) I can deduce that it's this line:

    return Long.valueOf(line).longValue();

The problem is line is a string of whitespace, not a numeric string. You can't expect to convert whitespace into a Long. That's why you get this error.

answered Nov 6, 2013 at 0:30

Daniel Kaplan

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Comments

MadProgrammer · Accepted Answer · 2013-11-06 00:43:31Z

0

I would have thought the exception self explanatory, " " can not be parsed as a Long.

Try using something like return line == null ? 0 : line.trim().isEmpty() ? 0 : Long.valueOf(line).longValue(); to determine the validity of the String value first and return a default value where it's not.

Should you not care about differentiating between a null String or an empty String you could also use return line == null || line.trim().isEmpty() ? 0 : Long.valueOf(line).longValue(); which may be easier to read

Or, throw some kind of exception where the value does not meet your exceptions if required

edited Nov 6, 2013 at 0:43

answered Nov 6, 2013 at 0:33

MadProgrammer

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2 Comments

Simon Arsenault Over a year ago

return line == null || line.trim().isEmpty() ? 0 : Long.valueOf(line).longValue(); would be better as it's easier to read.

MadProgrammer Over a year ago

@SimonArsenault No argument there, assuming that you wanted to treat a null and empty String the same way, just saying...

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