Scratch

You'll need to create a variable (counter) and (checking) for this.

set [counter v] to [1]
set [checking v] to [0]
if <[list v] contains [silver]>
repeat until <<(counter) < ((length of [list v])-(1))>or<(checking)=(1)>>
if <<item (counter) of [list v]>=[silver]>
delete (counter) of [list v]
set [checking v] to [1]
else
change [counter v] by (1)
end
end
end

TorbyFork234 wrote:

You'll need to create a variable (counter) and (checking) for this.

set [counter v] to [1]
set [checking v] to [0]
if <[list v] contains [silver]>
repeat until <<(counter) < ((length of [list v])-(1))>or<(checking)=(1)>>
if <<item (counter) of [list v]>=[silver]>
delete (counter) of [list v]
set [checking v] to [1]
else
change [counter v] by (1)
end
end
end

Pretty sure you need a "greater than" in that script, not a "less than"

BirdByte wrote:

Guys, you're over-complicating this.

set [counter v] to [1]
if <[list v] contains [silver]>
 repeat until <(item (counter) of [list v]) = [silver]>
  change [counter v] by (1)
 end
 delete (counter) of [list v]
end

That will only delete the first "silver" in the list. If you would like to delete all the "silver"s in the list, do this:

if <[list v] contains [silver]>
 repeat until <not <[list v] contains [silver]>>
  set [counter v] to [1]
  repeat until <(item (counter) of [list v]) = [silver]>
   change [counter v] by (1)
  end
  delete (counter) of [list v]
 end
end

That one is the simplest ^ and also easiest to understand. The counter variable goes through each item of the script, first, second etc until the item which corresponds to the counter value, eg the 34th value = silver. It then deletes this item.

The only problem with this is that you delete a value from the list. If the first and second items are silver, you would encounter a problem. Counter, =1, detects silver and deletes it, and then sets counter to 2. BUT, the original second item (silver) is NOW item 1, and the counter won't check it but instead move on. Theoretically, you could still have silver's in the list. To correct this simply don't change the counter if silver was detected:

if <[list v] contains [silver]>
set [counter v] to (1)
repeat until <(counter) > (length of [list v])>
if <(item (counter) of [list v]) = [silver]>
delete (counter) of [list v]
else
change [counter v] by (1)

Last edited by Prestige (2012-09-12 12:30:03)

MoreGamesNow wrote:

BirdByte wrote:

Guys, you're over-complicating this.
...code and stuff

if <[list v] contains [silver]>
 repeat until <not <[list v] contains [silver]>>
  set [counter v] to [1]
  repeat until <(item (counter) of [list v]) = [silver]>
   change [counter v] by (1)
  end
  delete (counter) of [list v]
 end
end

As long as we're simplifying, you don't even need that first if statement.     And using "repeat until" rather than an if statement means you are unnecessarily looping through parts of the list that you've already checked.

set [counter v] to (1)
repeat until <not <[list v] contains [silver]>>
if<(item (counter) of [list v]) = [silver]>
delete (counter) of [list v]
end
change [counter v] by (1)
end

That script won't work because counter will go higher than the length of the list. Use this, then:

repeat until <not <[list v] contains [silver]>>
 set [counter v] to (1)
 repeat (length of [list v])
  if<(item (counter) of [list v]) = [silver]>
   delete (counter) of [list v]
  end
  change [counter v] by (1)
 end
end

Last edited by BirdByte (2012-09-12 14:18:35)

sweet, they all worked. thx.

lol, so many options!