Unable to shorten strings

p5.js Coding Questions
1

This seems like a simple problem… but it has been driving me nuts!

var bob = '';

function setup() {
}
function draw() {
}

function keyTyped()
{
    bob += key;
    print(bob);
}
function keyPressed()
{
  if(keyCode == ENTER||keyCode == RETURN)
  {
    bob = '';
  }
  else if(keyCode == DELETE||keyCode == BACKSPACE) 
  {
    if(bob.length>0)
      bob = bob.substring(0,bob.length-1);
    print(bob.length);
  }  
}

Screenshot%20from%202018-10-02%2022-10-13
Screenshot from 2018-10-02 22-10-13.png1366×768 59.2 KB

1 Like
2

From this Java sketch link below: :coffee:
http://Studio.ProcessingTogether.com/sp/pad/export/ro.9Zo$UbIWYZEDR

void keyPressed() {
  if (key != CODED | idx < 0)  return;
  final int k = keyCode;
 
  final TextBox tbox = tboxes[idx];
  final int len = tbox.txt.length();
 
  if (k == LEFT)  tbox.txt = tbox.txt.substring(0, max(0, len-1));
  else if (k == RIGHT & len < tbox.lim-3)  tbox.txt += "    ";
}

Which in JS would be something like this: :upside_down_face:

function keyPressed() {
  if (idx < 0)  return;
  const k = keyCode;
 
  const tbox = tboxes[idx];
  const len = tbox.txt.length;
 
  if (k == LEFT)  tbox.txt = tbox.txt.substring(0, max(0, len-1));
  else if (k == RIGHT & len < tbox.lim-3)  tbox.txt += "    ";
}
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3

Thanks GTL, I got this idea after reading your suggestion.

function keyPressed()
{
  if(keyCode == ENTER||keyCode == RETURN)
    bob = '';
  else if(keyCode == DELETE||keyCode == BACKSPACE) 
  {
    if(bob.length>0)
      bob = bob.substring(0,bob.length-1);
  }
  else
    bob += key;
}
1 Like
4

yes, now with the if else structure
you not add the “Delete” key to the string ( before you do the delete ),
still what happens with all other keys what are not characters, like [SHIFT][x] ?

1 Like
5

You can also move in the bob.length > 0 condition expression to the main 1: :sunglasses:

else if (keyCode == DELETE | keyCode == BACKSPACE && bob.length)
  bob = bob.substring(0, bob.length - 1);
2 Likes
6

Thanks for the tips; I do plan to refactor later. I’m just in a mad dash to get the code running. I do like to reduce the number of lines… but I’ve found, in my limited javascript experience, that engines like extra brackets… not sure if it’s real or just perceived

7

The excerpt I’ve posted is 100% valid JS syntax! :money_mouth_face:

1 Like
8

GoToLoop, You’re right. It’s just perceived. Small correction though. The bob.length would need to be at the beginning.

else if (bob.length && keyCode == DELETE | keyCode == BACKSPACE )
  bob = bob.substring(0, bob.length - 1);

But it turns out that I don’t even need to test for the length because substring will accept -1 in the second argument without causing any issues I’ve seen.

kll, I’ve moved the code to keyTyped which helps.

function keyTyped()
{
  if(keyCode == ENTER||keyCode == RETURN) 
    bob = '';
  else if(keyCode == DELETE||keyCode == BACKSPACE)
     bob = bob.substring(0,bob.length-1);
  else
    bob += key;
  print(bob);
}
1 Like
9

keyTyped does not know keyCode but you can use it for
bob+=key;
AND [SHIFT][char] seems to work!
( but seems it adds DELETE key anyhow??)

and i would suggest a && ( b || c ) for the logic in keyPresssed()

else if (bob.length && (keyCode === DELETE || keyCode === BACKSPACE)) {    
 bob = bob.substring(0,bob.length-1); 
}

 print("bob "+bob+" length "+bob.length);
1 Like
10

Have you actually checked that out? AFAIK, both ways should result the same. :motorway:

Nice catch! JS’ String::substring() simply ignores invalid passed arguments to it: :crazy_face:

However, Java’s String::substring() will throw IndexOutOfBoundsException for any invalid arguments: :bug:

Docs.Oracle.com/en/java/javase/11/docs/api/java.base/java/lang/String.html#substring(int,int)

2 Likes