NumberFormatException occured
hi all
strPan = txtPan.getText();
Long iPAN = 0;
try {
iPAN = Long.parseLong(strPan.trim());
} catch(NumberFormatException e) {
String strErrMsg = "The PAN must be numeric";
JOptionPane.showMessageDialog(this, strErrMsg,"Error", JOptionPane.ERROR_MESSAGE);
return;
}
if(iPAN == 0 || iPAN >24) {
String strErrMsg = "The PAN must lie in the range 1 to 24";
JOptionPane.showMessageDialog(this, strErrMsg, "Error", JOptionPane.ERROR_MESSAGE);
return;
}
when i enter the PAN as 121212121212121212121212121212.
i got "The PAN must be numeric"
can any one help me what is the error.?
Well for starts you cannot convernt from int to long..... Long iPAN = 0;
Also 121212121212121212121212121212 isnt a long......
if passing in a correct long try the following
strPan = txtPan.getText();
long iPAN = 0; // updated your Long to my long
try {
> iPAN = Long.valueOf(strPan).longValue(); // converted differently....
} catch(NumberFormatException e) {
> hi all
>
> strPan = txtPan.getText();
> Long iPAN = 0;
> try {
> iPAN = Long.parseLong(strPan.trim());
> } catch(NumberFormatException e) {
> String strErrMsg = "The PAN must be
> numeric";
> JOptionPane.showMessageDialog(this, strErrMsg,
> ,
The following does not look like it will compile unless maybe autoboxing is allowing it:
iPAN = Long.parseLong(strPan.trim());
You should display strPan too be sure that it contains the string you expect.
Also, show us the actual stack trace with e.StackTrace().
jbisha at 2007-7-14 19:32:31 >
hi Tom_Timpsoneven this throws NumberFormatExceptionits there any other way. to convert string to long
hi jbisheven this is throws an exception for the strPan="121212121212121212121212121212"iPAN = Long.parseLong(strPan.trim());
> even this throws NumberFormatException
What String are you passing in? remember if you want to convert from a string to a long, the input MUST be a long.......
> its there any other way. to convert string to long
Yes long l = Long.valueOf(str).long.Value();
This will convert the input string "str" into it's long equivalence.
Is this any help?
> its there any other way. to convert string to long
Also :
// String s = "fred";// do this if you want an exception
String s = "100";
try {
long l = Long.parseLong(s.trim());
System.out.println("long l = " + l);
} catch (NumberFormatException nfe) {
System.out.println("NumberFormatException: " + nfe.getMessage());
}
OR
long l = Long.parseLong("123");
what am saying is
// String s = "fred";// do this if you want an exception
String s = "12121212121212121212";
try {
long l = Long.parseLong(s.trim());
System.out.println("long l = " + l);
} catch (NumberFormatException nfe) {
System.out.println("NumberFormatException: " + nfe.getMessage());
}
I believe the reason you are getting an exception is that your number is too big to fit into a long.From the JLS:For long, from -9223372036854775808 to 9223372036854775807, inclusive1212121212121212121212121212129223372036854775807
jbisha at 2007-7-14 19:32:31 >
The reason your getting number format exception is simpily becasue 12121212121212121212 is not a long.is there any other type you can use for iPan?
> 121212121212121212121212121212> 9223372036854775807Exactly..... there is no possible way of trying to do what you want...... unless you want to break the long down and add it all up again, but that might be a bit messy.
If you really need to check numbers this large, either check each character in the String to see if it is a number or use the BigInteger class: http://java.sun.com/j2se/1.5.0/docs/api/java/math/BigInteger.html
jbisha at 2007-7-14 19:32:31 >
> i enter the PAN as 121212121212121212121212121212.
> i got "The PAN must be numeric"
> can any one help me what is the error.?
PAN as in Primary Account Number (as with credit cards)?
Then stop trying to convert it. It should be manipulated as a string and not a numeric.
