I interpret your request like this: "Split a string at both CR and LF." which implies that CR+LF gives an empty string element. For instance, 'alpha'#13'beta'#10'gamma'#13#10'delta'
yields the five elements 'alpha'
, 'beta'
, 'gamma'
, ''
, and 'delta'
.
If so, and if you are using a non-ancient version of Delphi, this is really simple:
var S := 'alpha'#13'beta'#10'gamma'#13#10'delta';
var Parts := S.Split([#13, #10]);
for var Part in Parts do
ShowMessage(Part);
For old Delphi versions
The code above requires TStringHelper
(crucially) and also makes use of inline variable declarations, for in
loops, and generics.
For old Delphi versions, you can do it manually:
type
TStringArray = array of string;
function Split(const S: string): TStringArray;
var
Count: Integer;
const
Delta = 512;
procedure Add(const Part: string);
begin
if Length(Result) = Count then
SetLength(Result, Length(Result) + Delta);
Result[Count] := Part;
Inc(Count);
end;
var
p, i: Integer;
begin
Result := nil;
Count := 0;
p := 0; // previous delim
for i := 1 to Length(S) do
if S[i] in [#13, #10] then
begin
Add(Copy(S, Succ(p), i - p - 1));
p := i;
end;
Add(Copy(S, Succ(p)));
SetLength(Result, Count);
end;
procedure TForm1.FormCreate(Sender: TObject);
var
S: string;
Parts: TStringArray;
i: Integer;
begin
S := 'alpha'#13'beta'#10'gamma'#13#10'delta';
Parts := Split(S);
for i := 0 to High(Parts) do
ShowMessage(Parts[i]);
end;