List placements. Question.

[TessaAnnaMarie]

Okay i have 2 lists. I'll keep it simple here. 

list Numbers = ["1", "2", "3", "4", "5"]
list Letters = ["A", "B", "C", "D", "E"]

What I need to do is check to see if something is in a list. 

if(msg == llListFindList(Numbers, (Numbers)msg))
{
      llSay(0, **this is where I need help**);
}

Simple. But what I need the outcome to do is use the letter that corresponds to the number. Meaning if the message is "3", the outcome will say "C". I'm pretty sure this is doable, but I can't for the life of me figure out how. 

[CmpZ]

I'm not in front of a computer that can run SL at the moment, so I can't test it, but I believe you do this:
 

Do_The_Thing( string value ) {

    integer n;

    n = llListFindList( Numbers, [value] );

    if (-1 == n) {

          // not found

    } else {

        llOwnerSay( llList2String( Letters, n ) );

    }

}

The difference between this & your original code is that this one saves the result from llListFindList &, if it's in range (not -1), uses it as the index into the Letters list.

 

Edited June 8, 2023 by CmpZ
Corrected type of second actual argument to llListFindList

[TessaAnnaMarie]

17 minutes ago, CmpZ said:

I'm not in front of a computer that can run SL at the moment, so I can't test it, but I believe you do this:
 

Do_The_Thing( string value ) {

    integer n;

    n = llListFindList( Numbers, value );

    if (-1 == n) {

          // not found

    } else {

        llOwnerSay( llList2String( Letters, n ) );

    }

}

The difference between this & your original code is that this one saves the result from llListFindList &, if it's in range (not -1), uses it as the index into the Letters list.

 

Thank you! Some minor adjustments and yes, this works! Thank you!