Removing String

[robertjohns 0]

I have a string like

 

This is (string) which need to (remove)

 

How can I delete strings (string) and (remove) so that result should be

 

This is which need to

 

 

 

 

 

[Marsil 4]

  var s:= 'This is (string) which need to (remove)';
  var LeftIndex, RightIndex : Integer;

  LeftIndex := S.IndexOf (' (');
  while LeftIndex > -1 do
    begin
      RightIndex := S.IndexOf (')');
      if RightIndex = -1 then
        Break;
      S := S.Remove (LeftIndex, RightIndex - LeftIndex + 1);
      LeftIndex := S.IndexOf (' (');
    end;
    

 

Edited December 1, 2023 by Marsil
Changing code as question changed

[DelphiUdIT 176]

There are others variant of that function, with options to replace all occurrences and case sensitive or not.

 

See https://docwiki.embarcadero.com/Libraries/Athens/en/System.SysUtils.StringReplace for details.

[robertjohns 0]

7 hours ago, Marsil said:

  var s := 'This is (string) which need to (remove)';
  S := S.Replace (' (string)', '').Replace (' (remove)', '');
  Log ('S=' + s); // S=This is which need to

 

This is basic and you have tried to replace the string you already know.

 

String can be anything in brackets and need to delete the string including brackets .. all the string occurrences in line which is with bracket need to delet

[David Heffernan 2345]

Your specification in the original post is incomplete so it's not surprising that people don't know what you want. Your clarification is still unclear. 

 

Until you can define precisely what you want the code to do how could you expect anyone, even yourself, to be able to write it. 

[robertjohns 0]

34 minutes ago, David Heffernan said:

Your specification in the original post is incomplete so it's not surprising that people don't know what you want. Your clarification is still unclear. 

 

Until you can define precisely what you want the code to do how could you expect anyone, even yourself, to be able to write it. 

Thanks a lot for reply

 

I have Memo line in which there are such strings which are not specific otherwise I can use StringReplace method ,strings are in brackets () and I want to delete those strings which are with brackets

 

suppose

 

This is (string) which need to (remove)

 

Need to delete (string and (remove) from the above

 

[Leif Uneus 43]

In short, you want to remove all strings in brackets, including the brackets.

[mvanrijnen 123]

Use this: System.SysUtils.TStringHelper.IndexOf - RAD Studio API Documentation (embarcadero.com)

 

How to use the above to solve your problem is your job as a developer 😉

 

[robertjohns 0]

2 hours ago, Leif Uneus said:

In short, you want to remove all strings in brackets, including the brackets.

Yes right need to remove strings in brackets including brackets

[Marsil 4]

25 minutes ago, robertjohns said:

Yes right need to remove strings in brackets including brackets

As you changed your question I changed my answer above, I Hope it's what you are looking for.

BTW,  Your question has nothing to do with VCL, You should have posted it in General Help forum.

Edited December 1, 2023 by Marsil

[mvanrijnen 123]

17 minutes ago, Marsil said:

As you changed your question I changed my answer above, I Hope it's what you are looking for.

BTW,  Your question has nothing to do with VCL, You should have posted it in General Help forum.

and now try this ) with this ( text as it )() fill fail (sometimes) 😉

 

[Marsil 4]

11 minutes ago, mvanrijnen said:

and now try this ) with this ( text as it )() fill fail (sometimes) 😉

 

I don't understand what you just said!.

 

As the OP did not completely describe what the input data will be, The answer is not meant to be valid for any data input.

It's just a sample describing how to remove occurrences from a string. Can be adapter by OP to his own needs as he wish.

If you have not noticed I didn't write this as ready to use function , Just a sample code snippet.

Edited December 1, 2023 by Marsil

[Serge_G 87]

Did you consider using regular expressions ?

uses system.regularexpressions;

procedure TForm1.Button1Click(Sender: TObject);
var reg : TRegEx;
    r : String;
const s = 'This is (string) which need to (remove)';
begin
reg:=TRegEx.Create('\([^()]*\)');
r:=reg.Replace(s,'');
showmessage(r);
end;

 

Edited December 2, 2023 by Serge_G
added, code sample

[Lars Fosdal 1792]

[David Heffernan 2345]

On 12/1/2023 at 5:01 AM, robertjohns said:

This is (string which need to (remove)

What should happen here

[JohnLM 14]

Quote

This is (string) which (string) need to (remove)

..or here.

[Lars Fosdal 1792]

My interpretation:
This is (string which need to (remove) -> This is

This is (string) which (string) need to (remove) -> This is which need to

[JohnLM 14]

Serge_G's routine works, including when multiples of, i.e., (string) .. (string) 

 

But it does not remove the extra spaces. And Trim() does not remove the spaces in-between char/strings unless the original string includes spaces at the first and last position, i.e., 

 

Trim('  This is (string) which (string) need to (remove)  ')  ---  works

Trim('This is (string)   which (string) need   to (remove)')  ---  fails 

 

So, the routine would require more work/additional code.

 

Edited December 3, 2023 by JohnLM

[Uwe Raabe 2057]

I would split removing the bracketed strings and removing double blanks in separate methods.

 

I am assuming the strings are well-formed here, which means that each opening bracket matches a closing bracket, no orphaned brackets and no nesting. Otherwise use one of the other algorithms shown above.

 

function RemoveInBrackets(const AString: string): string;
begin
  var arr := AString.Split(['(', ')']);
  for var I := High(arr) downto 0 do
  begin
    if Odd(I) then
      Delete(arr, I, 1);
  end;
  Result := string.Join('', arr);
end;

function RemoveDoubleBlanks(const AString: string): string;
begin
  Result := string.Join(' ', AString.Split([' '], TStringSplitOptions.ExcludeEmpty));
end;

 

 

[JohnLM 14]

Here is my version of a (string) removal for this practice: 

 

function StrRemove(s: string; StrToRemove:string): string;
begin
  while pos(strToRemove,s,1) <> 0 do begin // recursive
    if pos(strToRemove, s,1)  > 0 then delete(s,pos(strToRemove,s,1),length(strToRemove));
  end;
  result := s;
end;

// call with 

var s: string; 
begin
  s := 'This is (string) which (string) need to (remove).'; 
  s := StrRemove(s,'(string)');
  s := StrRemove(s,'(remove)');
end; 

 

Edited December 3, 2023 by JohnLM
updated code snippet - removed Trim(), it does not work in this routine.

[limelect 48]

Use RxStrUtils.pas large amount of functions that do exactly that

Why invent the weel?

Edited December 6, 2023 by limelect

[Tom F 83]

@Lars Fosdal  Ha!  I actually literally LOL-ed on that.  

 

ChatGPT has changed my relationship to RegEx. As we know, it's not always correct, but, hey, what a great tool!

 

 

 

[pmcgee 10]

The regex solution above has an issue with nested parentheses ... but could be repeated.

[pmcgee 10]

Well. I learnt about something today. Recursive regexes.

Edited December 7, 2023 by pmcgee

[mvanrijnen 123]

49 minutes ago, pmcgee said:

Well. I learnt about something today. Recursive regexes.

double trouble 🙂