java.lang.NumberFormatException: For input string: "10.0"

java exception numberformatexception

George · Oct 10, 2013 · Viewed 20.5k times · Source

This code must validate input data from the findActions() method:

try {
    System.out.println(findActions(lookingArea.substring(0, right)));// always printing valid number string
    Integer.parseInt(findActions(lookingArea.substring(0, right)));// checking for number format
}
catch(NumberFormatException exc) {
    System.out.println(exc);
}

But I always have java.lang.NumberFormatException: For input string: "*number*" that is so strange, because checking with System.out.println(findActions(lookingArea.substring(0, right)));,

I get *number* like 10.0

Answer

Integer.parseInt doesn't expect the . character. If you're sure it can be converted to an int, then do one of the following:

Eliminate the ".0" off the end of the string before parsing it, or
Call Double.parseDouble, and cast the result to int.

Quoting the linked Javadocs above:

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.

java.lang.NumberFormatException: For input string: "10.0"

Answer

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