SubStringLastIndex

Summary

User-Defined Function: integer uSubStringLastIndex( string vStrSrc, string vStrTst );

Returns a integer that is the positive index of the last vStrTst within vStrSrc

• vStrSrc: source string to check
• vStrTst: string to look for

if vStrTst is not found in vStrSrc -1 is returned.
the index of the first character is 0

Code:

• LSO: 182 bytes
• MONO: 1024 bytes

integer uSubStringLastIndex( string vStrSrc, string vStrTst ){
    integer vIdxFnd =
      llStringLength( vStrSrc ) -
      llStringLength( vStrTst ) -
      llStringLength(
        llList2String(
          llParseStringKeepNulls( vStrSrc, (list)vStrTst, [] ),
          0xFFFFFFFF ) //-- (-1)
        );
    return (vIdxFnd | (vIdxFnd >> 31));
}
/*//--                       Anti-License Text                         --//*/
/*//     Contributed Freely to the Public Domain without limitation.     //*/
/*//   2009 (CC0) [ http://creativecommons.org/publicdomain/zero/1.0 ]   //*/
/*//  Void Singer [ https://wiki.secondlife.com/wiki/User:Void_Singer ]  //*/
/*//--                                                                 --//*/

Code: Alternative implementation - may be faster, tested.

(warning: uses recursion, high potential for LSL stack heap collision when processing large and/or highly repetitive strings)

integer uSubStringLastIndex(string vStrSrc, string vStrTst)
{
    if (vStrTst == "")
        return 0;
    integer index = llSubStringIndex(vStrSrc, vStrTst);
    integer index2;
    if (index != -1) //found, look again
        index2 = uSubStringLastIndex(llGetSubString(vStrSrc, index + 1, -1), vStrTst) + 1;
    return index + index2;
}
/*// Contributed Freely to the Public Domain without limitation by Sasun Steinbeck. //*/

Third solution, tested

integer uSubStringLastIndex(string hay, string pin)
{
    integer index2 = -1;
    integer index;
    if (pin == "")    
        return 0;
    while (~index)
    {
        index = llSubStringIndex( llGetSubString(hay, ++index2, -1), pin);
        index2 += index;
    }
    return index2;
}

Fourth solution, tested

integer uSubStringLastIndex(string haystack, string needle)
{
    return llSubStringIndex(llReplaceSubString(haystack, needle, NAK, -1), NAK); //make sure the value for NAK won't ever natively/naturally occur in source string, else, use a different unique string for pattern
}
/*// Contributed Freely to the Public Domain without limitation by Lucia Nightfire. //*/

Caveats

• Performs a literal match (case sensitive).
  • Wildcards and RegEx are not supported.
• Attempting to match an empty string ("") will return 0 instead of -1.

Notes

• This function is operates exactly like llSubStringIndex (including caveats), from the opposite end of the string.