Best way to simplify this code, more efficient

Question

My question is, is there a way to make this code more efficient or write it in a simple way? javascript by the way.

switch (tempvar1) {
  case 1:
    currentSlide = 'slide1';
    showaslide('ppslide1');
    break;
  case 2:
    currentSlide = 'slide2';
    showaslide('ppslide2');
    break;
  case 3:
    currentSlide = 'slide3';
    showaslide('ppslide3');
    break;
  case 4:
    currentSlide = 'slide4';
    showaslide('ppslide4');
    break;
  case 5:
    currentSlide = 'slide5';
    showaslide('ppslide5');
    break;
  case 6:
    currentSlide = 'slide6';
    showaslide('ppslide6');
    break;
  // 20 total cases
}

Answer 1

currentSlide = 'slide' + tempvar1;
showaslide('ppslide' + tempvar1);

Answer 2

How about:

currentSlide = 'slide' + tempvar1;
showaslide('ppslide' + tempvar1);

Answer 3

This can be done much more elegant:

if (tempvar1 <= 20) {
  currentSlide = 'slide' + tempvar1;
  showaslide('ppslide' + tempvar1);
}

In JavaScript, integer values are automatically converted to strings if combined with them. The + operator concatenates the strings.

EDIT: Followup on your comment question to check if tempvar is an integer. One possiblity to check this (there are also other modifications to round up the code):

if (!isNaN(parseInt(tempvar1))) {
  if (tempvar1 >= 1) && (tempvar1 <= 20) {
    currentSlide = 'slide' + tempvar1;
    showaslide('ppslide' + tempvar1);
  } else {
    // Some error handling or default case here if needed
    // thx to eyelidlessness ;)
  }
}

The first if statement tries to treat tempvar1 as an integer value and only continues if this works.

Answer 4

You don't need the switch statement:

currentSlide = 'slide' + tempvar1;
showaslide('ppslide' + tempvar1);