CODEWITHSUNDEEP
javatype-conversionparseintnumberformatexception
Kent Godfrey
Kent Godfrey

Reputation: 73

Converting string into a readable int from a JLabel

I need to apply an interest method to a number in a JLabel. I can manage to do it from a Jtextfield, but for some reason I cannot get it to work on the JLabel.

Here is the code that is initiated when the Jbutton is pressed:

private void jButton4ActionPerformed(java.awt.event.ActionEvent evt) {                                         
    interest(Integer.parseInt(balanceLabel.getText()));

balanceLabel is the name of the label I am trying to work with.

Here is the error that is returned when I press the button:

Exception in thread "AWT-EventQueue-0" java.lang.NumberFormatException: For input string: "£1000.0"
at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65)

I have researched the problem and it seems it is extremely common but for some reason I cannot apply other answers to my situation as I lack the knowledge to do so.

Upvotes: 1

Views: 2287

Answers (2)

XaolingBao
XaolingBao

Reputation: 1044

Well, from what I can see you are trying to convert '£' which isn't an integer, so you are getting an exception for that. You also have a decimal value, which an integer cannot handle, so the exception is also being thrown for that reason.

public static int parseInt(String s) throws NumberFormatException

Parses the string argument as a signed decimal integer. The characters in the string must all be decimal digits, except that the first character may be an ASCII minus sign '-' ('\u002D') to indicate a negative value or an ASCII plus sign '+' ('\u002B') to indicate a positive value. The resulting integer value is returned, exactly as if the argument and the radix 10 were given as arguments to the parseInt(java.lang.String, int) method.

Parameters: s - a String containing the int representation to be parsed

Returns: the integer value represented by the argument in decimal.

Throws: NumberFormatException - if the string does not contain a parsable integer.

what you could do, if you know that everything will be '£xxxx.xx," is you can change this line

  interest(Integer.parseInt(balanceLabel.getText()));

to this

interest(Double.parseDouble(balanceLabel.getText().substring(1));

which would then return "1000.0"

public String substring(int beginIndex)

Returns a string that is a substring of this string. The substring begins with the character at the specified index and extends to the end of this string. Examples:

"unhappy".substring(2) returns "happy"

"Harbison".substring(3) returns "bison"

"emptiness".substring(9) returns "" (an empty string)

Parameters: beginIndex - the beginning index, inclusive.

Returns: the specified substring.

Throws: IndexOutOfBoundsException - if beginIndex is negative or larger than the length of this String object.

Upvotes: 1

dambros
dambros

Reputation: 4392

The problem is the £ AND the . since you are trying an int conversion.

Use a float instead:

Float.parseFloat(balanceLabel.getText().substring(1));

This way you can have decimal values, which makes sense for currency.

Upvotes: 2

Related Questions