RegEx Help, Please

[PepsiHog]

Hello Everyone,

Ok I have a list of numbers separated by a semi-colon. Such as:
i.e. 1;2;3;4;5;6;7;8;9;1;2;3;23;34;51;

In the above list 1;2;3; repeats. The following RegEx removes repeats.

RegEx>\b(\d\d?);(?!.*\b\1\b),StringInQuestion,0,match,NumbersExcluded,1,,RepeatsRemoved

This results in:
i.e. 4;5;6;7;8;9;1;2;3;23;34;51;

The 1;2;3; At the beginning was removed. But, and this is where I need some help, 1;2;3; still remains in the string.

I want both instances of any numbers that repeat to be removed. So the desired result would be:
i.e. 4;5;6;7;8;9;23;34;51;

Both instances of 1;2;3; have been removed in the last above example.

How could I accomplish this with RegEx?

[Grovkillen]

Single RegEx will probably not do exactly what you want (memory for pulling of that one might be big?). This is how I would do it:

Code: Select all

Let>STRING=;1;2;3;4;5;6;7;8;9;1;2;3;23;34;51;
Let>REGEX_PATTERN=\b(\d+)\b(?=.*\b\1\b)
RegEx>REGEX_PATTERN,STRING,0,FOUND,FOUND_count,0,,
If>FOUND_count>0
  Let>k=0
  Repeat>k
    Let>k=k+1
    Let>TEMP_to_be_removed=FOUND_%k%
    StringReplace>STRING,;%TEMP_to_be_removed%;,;,STRING
  Until>k=FOUND_count
Endif>

I put the semi colon in the beginning of the string to simplify the replacement of the duplicates.

Since this one will match all duplicates, your loop will be iterated more times than needed, you could have a variable that stored already removed duplicates and thus speed the loop up.