java.lang.NumberFormatException：对于输入字符串：“”
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原文地址: http://stackoverflow.com/questions/19801823/
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java.lang.NumberFormatException: For input string: " "
提问by
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 *
*运行项目“线程“main”中的异常java.lang.NumberFormatException：对于输入字符串：“我的类的功能是避免重新运行应用程序时出现下一个错误。可以帮助定位故障。非常感谢您 *
here is my class lexec
这是我的班级 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);
}
}
}
采纳答案by Daniel Kaplan
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:
虽然你没有告诉我们哪一行给了我们错误（即使你应该有）我可以推断出这是这一行：
return Long.valueOf(line).longValue();
The problem is lineis 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.
问题是line一串空格，而不是一个数字串。您不能期望将空格转换为Long. 这就是您收到此错误的原因。
回答by MadProgrammer
I would have thought the exception self explanatory, " "can not be parsed as a Long.
我原以为异常是不言自明的，" "不能解析为Long.
Try using something like return line == null ? 0 : line.trim().isEmpty() ? 0 : Long.valueOf(line).longValue();to determine the validity of the Stringvalue first and return a default value where it's not.
尝试使用类似的方法return line == null ? 0 : line.trim().isEmpty() ? 0 : Long.valueOf(line).longValue();首先确定String值的有效性，然后在不存在的情况下返回默认值。
Should you not care about differentiating between a nullStringor an empty Stringyou could also use return line == null || line.trim().isEmpty() ? 0 : Long.valueOf(line).longValue();which may be easier to read
如果您不关心区分 anullString或空，String您也可以使用return line == null || line.trim().isEmpty() ? 0 : Long.valueOf(line).longValue();它可能更容易阅读
Or, throw some kind of exception where the value does not meet your exceptions if required
或者，如果需要，在值不符合您的异常的情况下抛出某种异常
