Scratch

You can try something like this:

when gf clicked
set [counter v] to [0]
repeat (length of [list v])
change [counter v] by (1)
if <(item (counter) of [list v]) = [whatever you want to remove]>
delete (counter) of [list v]
end
end

jvvg wrote:

You can try something like this:

when gf clicked
set [counter v] to [0]
repeat (length of [list v])
change [counter v] by (1)
if <(item (counter) of [list v]) = [whatever you want to remove]>
delete (counter) of [list v]
end
end

Actually, that wouldn't work 100% of the time.

For example, lets say that we want to remove "pie" from the list and we have this list:

candy
pie
pie
chocolate

As soon as your script removes the first "pie", it will move onto the "chocolate" and entirely skip the second "pie", so this is the script I would use:

when gf clicked
set [counter v] to (1)
repeat until <not <[list v] contains [whatever you want to remove]>>
if <(item (counter) of [list v]) = [whatever you want to remove]>
delete (counter) of [list v]
else
change [counter v] by (1)
end
end

ImagineIt wrote:

ErnieParke wrote:

jvvg wrote:

You can try something like this:

when gf clicked
set [counter v] to [0]
repeat (length of [list v])
change [counter v] by (1)
if <(item (counter) of [list v]) = [whatever you want to remove]>
delete (counter) of [list v]
end
end

Actually, that wouldn't work 100% of the time.

For example, lets say that we want to remove "pie" from the list and we have this list:

candy
pie
pie
chocolate

As soon as your script removes the first "pie", it will move onto the "chocolate" and entirely skip the second "pie", so this is the script I would use:

when gf clicked
set [counter v] to (1)
repeat until <not <[list v] contains [whatever you want to remove]>>
if <(item (counter) of [list v]) = [whatever you want to remove]>
delete (counter) of [list v]
else
change [counter v] by (1)
end
end

Smart.

Thanks you.

Edit: Wait, shouldn't this be in Help with Scripts and not Questions about Scratch?

Last edited by ErnieParke (2012-09-30 15:56:59)