I'm having an issue with a java program trying to get a string input from a joptionpane menu with a prompt box. With returning a string input. I don't know if im going about it all wrong by trying to use
String.parseString(input)
Im very much a beginner with this so any help would have to be as simple as possible or a correction outright.
private static String getStringInput (String prompt) {
String input = EZJ.getUserInput(prompt);
return String.parseString(input);
}
UseCalls.java:27: error: cannot find symbol
return String.parseString(input);
^
symbol: method parseString(String)
location: class String
1 error
Here is a sample of the menu Im trying to use it with
do {
userInput = mainMenu();
if (userInput.equals("1")) {
String name = getStringInput("Name?");
String address = getStringInput("Address?");
call[numCalls++] = new Call();
}
} while (!userInput.equals("0"));
}
Here is the EZJ mini method
public class EZJ {
public static String getUserInput (String prompt) {
return JOptionPane.showInputDialog(prompt);
}
public static void dialog(String inputValue) {
JOptionPane.showMessageDialog ( null, inputValue );
}
}
3 Answers
You don't need to parse the string, it's defined as a string already.
Just do:
private static String getStringInput (String prompt) {
String input = EZJ.getUserInput(prompt);
return input;
}
As you see in an error UseCalls.java:27: error: cannot find symbol
return String.parseString(input); there is no method parseString in String class. There is no need to parse it as long as JOptionPane.showInputDialog(prompt); already returns a string.
If you're really bent upon converting Integer to String value, I suggest use String.valueOf(YourIntegerVariable). More details can be found at:
