[SOLVED]Using TRegExpr with a list, mixing a numbers list and a range of numbers

[devEric69]

Hello,
I would like to check the entry of a user with class TRegExpr.

He should only be able to capture a mixture of 2 constraints:
1: a list of numbers between 1 and 15. For example: 1,5,7,11
2: a range of numbers, for example [4-15], which means 4..15, ie 4,5,6,7,8,9,10,11,12,13,14,15.

And he could mix the two constraints together. For example: 1,[3-13],15. Of course, knowing how to do 1°) and 2°), it'll be a simple OR like this: ((1)*|(2)*)

Any leads?
Regards.

[devEric69]

1°) seems to me finally feasible, with '(\d{1,2})', and then by converting the regex's exactions into integer, and then checking if these integers are between the minimum allowed value and the maximum allowed value.

I always turn around how to make the range of numbers in 2°).

[Roland57]

1°) seems to me finally feasible, with '(\d{1,2})', and then by converting the regex's exactions into integer, and then checking if these integers are between the minimum allowed value and the maximum allowed value.

I always turn around how to make the range of numbers in 2°).

Hello! Here is my proposition. As you can see, I had the same idea as you. 

Code: Pascal  [Select][+][-]

1. uses
2.   SysUtils, Classes, RegExpr;
3.  
4. function Check(const AStr: string): boolean;
5. var
6.   E, F: TRegExpr;
7.   L: TStringList;
8.   I, J, K: integer;
9. begin
10.   result := TRUE;
11.   E := TRegExpr.Create('^\d{1,2}$');
12.   F := TRegExpr.Create('^\[(\d{1,2})-(\d{1,2})\]$');
13.   L := TStringList.Create;
14.   L.DelimitedText := AStr;
15.   try
16.     for I := 0 to L.Count - 1 do
17.     begin
18.       if E.Exec(L[I]) then
19.       begin
20.         J := StrToIntDef(L[I], 0);
21.         if (J >= 1) and (J <= 15) then
22.         begin
23.           WriteLn('OK: ', L[I], ' (J=', J, ')');
24.           Continue;
25.         end;
26.       end;
27.       if F.Exec(L[I]) then
28.       begin
29.         J := StrToIntDef(F.Match[1], 0);
30.         K := StrToIntDef(F.Match[2], 0);
31.         if (J >= 1) and (J <= 15) and (K >= 1) and (K <= 15) then
32.         begin
33.           WriteLn('OK: ', L[I], ' (J=', J, '; K=', K, ')');
34.           Continue;
35.         end;
36.       end;
37.       WriteLn('NOK: ', L[I]);
38.       result := FALSE;
39.       Break;
40.     end;
41.   finally
42.     E.Free;
43.     F.Free;
44.     L.Free;
45.   end;
46. end;
47.  
48. var
49.   S: string;
50.  
51. begin
52.   S := '1,[3-13],15';
53.   WriteLn(Check(S));
54. end.
55.  

OK: 1 (J=1)
OK: [3-13] (J=3; K=13)
OK: 15 (J=15)
TRUE

[devEric69]

Hello @Roland57.

I had gone in another direction with a beginning of regex like this: '([0-1][0-9]|[0-9])\-([0-1][0-9]|[0-9])' based on https://www.regular-expressions.info/numericranges.html. It must also be able to do it, too.

But anyway, thank you, because your solution is fine with me .