Input an integer value

xristosk

Seems dump but I cannot find a way to input an integer using the keyboard. I tried the following but it doesn't work. int a; if ((keyPressed == true) { a=key; }

KevinWorkman

What exactly do you mean when you say it doesn't work?

Keep in mind that the key variable is a char, which holds the unicode character. A unicode character is just an integer that represents a specific character. 'A' is 65, 'B' is 66, etc.

In other words, if the user presses the 2 key, key will be the char '2'.

The unicode value of the '2' character is 50, which is what you're setting i equal to.

You need a way to map that 49 back to the digit 2. One way to do that is by using the Character class:

a = Character.digit(key, 10);

If you have more questions, please post your code in the form of an MCVE, and don't forget to use the code button to preserve formatting.

GoToLoop

http://forum.processing.org/two/discussion/8045/how-to-format-code-and-text

• Variables declared in a locally scope cease to exist once that scope ends.
• If you need other places to see variable a, it needs to be declared globally as a field.
• Also we don't need to test for true b/c it can be safely omitted:
  if (keyPressed) a = key;

xristosk

Thank you for your response. I understand that key is of char type. I want my program to read an integer value in the setup function. Do I have to use a = Character.digit(key, 10); or I could use a more simple instruction?

KevinWorkman

What do you mean by more simple? That's a one-line solution. You could probably get away with using subtraction, but that seems kludgy to me.

There are probably hundreds of ways to do this, and google is your friend.

GoToLoop

... or I could use a more simple instruction?

Alternative implementation of Character.digit():

static final int digitToNum(char ch) {
  int num = ch - '0';
  return num >= 0 & num <= 9? num : -1;
}

KevinWorkman

@GoToLoop That code doesn't use your num variable, so it's not returning the correct value. Please run example code before posting it.

GoToLoop

Oops! Forgot to change ch to num within the return as its output! Fixed now! X_X

hamoid

This shows a dialog asking for a number:

import javax.swing.JOptionPane;

int i = 0;
String r = JOptionPane.showInputDialog(null, "What number?", "Decide", JOptionPane.QUESTION_MESSAGE);
try {
  i = Integer.parseInt(r);
} catch(NumberFormatException e) {
  println("you did not enter a number!");
}

println("i =", i);

xristosk

@hamoid Grateful for your post, this is what I want, but i get an error message.

Sinchai

Input to string. 
Converse string to int.
in setup
noLoop()

 in draw() 
check key  0...9
str += key
check ENTER to stop input
convers to int
clear str

in keyPressed()
redraw()

GoToLoop

It shoulda worked for JOptionPane belongs to standard Java API! :-/
Or are you under Android's modified Java version (Dalvik) by chance? :-?
Nonetheless, a similar alternative for @hamoid's version: \m/

// forum.processing.org/two/discussion/9563/input-an-integer-value

import static javax.swing.JOptionPane.*;

String input = showInputDialog("Type in integer number...");
int num = parseInt(input == null? "" : input, MIN_INT);

if (input == null)  showMessageDialog(null, "You didn't enter anything!", "Alert", ERROR_MESSAGE);
else if (num == MIN_INT)  showMessageDialog(null, "Entry \"" + input + "\" isn't a number!", "Alert", ERROR_MESSAGE);
else showMessageDialog(null, "Number " + num + " has been registered.", "Info", INFORMATION_MESSAGE);

exit();

hamoid

@xristosk I wonder if the Processing version makes a difference. It worked for me with 2.2.1 and 3.0a5. I no longer have 1.5.1.

Something in the direction of what @Sinchai shared would work: build your own input box that accepts numbers, enter, maybe backspace.

GoToLoop

Online textbox input: O:-)
http://studio.processingtogether.com/sp/pad/export/ro.9Zo$UbIWYZEDR/latest

xristosk

I can't run 2.2.1 so I still use 1.5.1. It is a java problem I haven't resolve. But I will test the code on a newer Processing version.

GoToLoop

• @xristosk, if you're using a version that's too old or it's alpha/beta you shoulda been upfront about that! :-w
• That would help us give you more fitting answers! :-<
• Nonetheless, even Processing 1.5.1 shouldn't have any problems w/ standard Java API than any other Processing version! 8-X
• My own JOptionPane example works in my Processing 1.5.1 as flawless as the v2.2.1!
• Just make sure to remove/rename Processing 1.5.1's "/java/" subfolder.
• Doing so forces v1.5.1 to use system's own installed Java!
• Processing v1.5.1 works even w/ latest 64-bit JDK 8!

xristosk

OK, it worked in Processing 2.2.1 Thank you all. I 'm grateful...