Split a string into an array of strings based on a delimiter - delphi

Question

I'm trying to find a Delphi function that will split an input string into an array of strings based on a delimiter. I've found a lot from searching the web, but all seem to have their own issues and I haven't been able to get any of them to work.
I just need to split a string like:
"word:doc,txt,docx" into an array based on ':'. The result would be
['word', 'doc,txt,docx']. How can I do that?

Answer 1

you can use the TStrings.DelimitedText property for split an string
check this sample
program Project28;
{$APPTYPE CONSOLE}
uses
Classes,
SysUtils;
procedure Split(Delimiter: Char; Str: string; ListOfStrings: TStrings) ;
begin
ListOfStrings.Clear;
ListOfStrings.Delimiter := Delimiter;
ListOfStrings.StrictDelimiter := True; // Requires D2006 or newer.
ListOfStrings.DelimitedText := Str;
end;
var
OutPutList: TStringList;
begin
OutPutList := TStringList.Create;
try
Split(':', 'word:doc,txt,docx', OutPutList) ;
Writeln(OutPutList.Text);
Readln;
finally
OutPutList.Free;
end;
end.
UPDATE
See this link for an explanation of StrictDelimiter.

Answer 2

There is no need for engineering a Split function. It already exists, see: Classes.ExtractStrings.
Use it in a following manner:
program Project1;
{$APPTYPE CONSOLE}
uses
Classes;
var
List: TStrings;
begin
List := TStringList.Create;
try
ExtractStrings([':'], [], PChar('word:doc,txt,docx'), List);
WriteLn(List.Text);
ReadLn;
finally
List.Free;
end;
end.
And to answer the question fully; List represents the desired array with the elements:
List[0] = 'word'
List[1] = 'doc,txt,docx'

Answer 3

You can use StrUtils.SplitString.
function SplitString(const S, Delimiters: string): TStringDynArray;
Its description from the documentation:
Splits a string into different parts delimited by the specified
delimiter characters.
SplitString splits a string into different parts delimited by the specified delimiter characters. S is the string to be split.
Delimiters is a string containing the characters defined as delimiters.
SplitString returns an array of strings of type System.Types.TStringDynArray that contains the split parts of the
original string.

Answer 4

Using the SysUtils.TStringHelper.Split function, introduced in Delphi XE3:
var
MyString: String;
Splitted: TArray<String>;
begin
MyString := 'word:doc,txt,docx';
Splitted := MyString.Split([':']);
end.
This will split a string with a given delimiter into an array of strings.

Answer 5

I always use something similar to this:
Uses
StrUtils, Classes;
Var
Str, Delimiter : String;
begin
// Str is the input string, Delimiter is the delimiter
With TStringList.Create Do
try
Text := ReplaceText(S,Delim,#13#10);
// From here on and until "finally", your desired result strings are
// in strings[0].. strings[Count-1)
finally
Free; //Clean everything up, and liberate your memory ;-)
end;
end;

Answer 6

Similar to the Explode() function offered by Mef, but with a couple of differences (one of which I consider a bug fix):
type
TArrayOfString = array of String;
function SplitString(const aSeparator, aString: String; aMax: Integer = 0): TArrayOfString;
var
i, strt, cnt: Integer;
sepLen: Integer;
procedure AddString(aEnd: Integer = -1);
var
endPos: Integer;
begin
if (aEnd = -1) then
endPos := i
else
endPos := aEnd + 1;
if (strt < endPos) then
result[cnt] := Copy(aString, strt, endPos - strt)
else
result[cnt] := '';
Inc(cnt);
end;
begin
if (aString = '') or (aMax < 0) then
begin
SetLength(result, 0);
EXIT;
end;
if (aSeparator = '') then
begin
SetLength(result, 1);
result[0] := aString;
EXIT;
end;
sepLen := Length(aSeparator);
SetLength(result, (Length(aString) div sepLen) + 1);
i := 1;
strt := i;
cnt := 0;
while (i <= (Length(aString)- sepLen + 1)) do
begin
if (aString[i] = aSeparator[1]) then
if (Copy(aString, i, sepLen) = aSeparator) then
begin
AddString;
if (cnt = aMax) then
begin
SetLength(result, cnt);
EXIT;
end;
Inc(i, sepLen - 1);
strt := i + 1;
end;
Inc(i);
end;
AddString(Length(aString));
SetLength(result, cnt);
end;
Differences:
aMax parameter limits the number of strings to be returned
If the input string is terminated by a separator then a nominal "empty" final string is deemed to exist
Examples:
SplitString(':', 'abc') returns : result[0] = abc
SplitString(':', 'a:b:c:') returns : result[0] = a
result[1] = b
result[2] = c
result[3] = <empty string>
SplitString(':', 'a:b:c:', 2) returns: result[0] = a
result[1] = b
It is the trailing separator and notional "empty final element" that I consider the bug fix.
I also incorporated the memory allocation change I suggested, with refinement (I mistakenly suggested the input string might at most contain 50% separators, but it could conceivably of course consist of 100% separator strings, yielding an array of empty elements!)

Answer 7

Explode is very high speed function, source alhoritm get from TStrings component.
I use next test for explode:
Explode 134217733 bytes of data, i get 19173962 elements, time of work: 2984 ms.
Implode is very low speed function, but i write it easy.
{ ****************************************************************************** }
{ Explode/Implode (String <> String array) }
{ ****************************************************************************** }
function Explode(S: String; Delimiter: Char): Strings; overload;
var I, C: Integer; P, P1: PChar;
begin
SetLength(Result, 0);
if Length(S) = 0 then Exit;
P:=PChar(S+Delimiter); C:=0;
while P^ <> #0 do begin
P1:=P;
while (P^ <> Delimiter) do P:=CharNext(P);
Inc(C);
while P^ in [#1..' '] do P:=CharNext(P);
if P^ = Delimiter then begin
repeat
P:=CharNext(P);
until not (P^ in [#1..' ']);
end;
end;
SetLength(Result, C);
P:=PChar(S+Delimiter); I:=-1;
while P^ <> #0 do begin
P1:=P;
while (P^ <> Delimiter) do P:=CharNext(P);
Inc(I); SetString(Result[I], P1, P-P1);
while P^ in [#1..' '] do P:=CharNext(P);
if P^ = Delimiter then begin
repeat
P:=CharNext(P);
until not (P^ in [#1..' ']);
end;
end;
end;
function Explode(S: String; Delimiter: Char; Index: Integer): String; overload;
var I: Integer; P, P1: PChar;
begin
if Length(S) = 0 then Exit;
P:=PChar(S+Delimiter); I:=1;
while P^ <> #0 do begin
P1:=P;
while (P^ <> Delimiter) do P:=CharNext(P);
SetString(Result, P1, P-P1);
if (I <> Index) then Inc(I) else begin
SetString(Result, P1, P-P1); Exit;
end;
while P^ in [#1..' '] do P:=CharNext(P);
if P^ = Delimiter then begin
repeat
P:=CharNext(P);
until not (P^ in [#1..' ']);
end;
end;
end;
function Implode(S: Strings; Delimiter: Char): String;
var iCount: Integer;
begin
Result:='';
if (Length(S) = 0) then Exit;
for iCount:=0 to Length(S)-1 do
Result:=Result+S[iCount]+Delimiter;
System.Delete(Result, Length(Result), 1);
end;

Answer 8

var
su : string; // What we want split
si : TStringList; // Result of splitting
Delimiter : string;
...
Delimiter := ';';
si.Text := ReplaceStr(su, Delimiter, #13#10);
Lines in si list will contain splitted strings.

Answer 9

You can make your own function which returns TArray of string:
function mySplit(input: string): TArray<string>;
var
delimiterSet: array [0 .. 0] of char;
// split works with char array, not a single char
begin
delimiterSet[0] := '&'; // some character
result := input.Split(delimiterSet);
end;

Answer 10

Here is an implementation of an explode function which is available in many other programming languages as a standard function:
type
TStringDynArray = array of String;
function Explode(const Separator, S: string; Limit: Integer = 0): TStringDynArray;
var
SepLen: Integer;
F, P: PChar;
ALen, Index: Integer;
begin
SetLength(Result, 0);
if (S = '') or (Limit < 0) then Exit;
if Separator = '' then
begin
SetLength(Result, 1);
Result[0] := S;
Exit;
end;
SepLen := Length(Separator);
ALen := Limit;
SetLength(Result, ALen);
Index := 0;
P := PChar(S);
while P^ <> #0 do
begin
F := P;
P := AnsiStrPos(P, PChar(Separator));
if (P = nil) or ((Limit > 0) and (Index = Limit - 1)) then P := StrEnd(F);
if Index >= ALen then
begin
Inc(ALen, 5);
SetLength(Result, ALen);
end;
SetString(Result[Index], F, P - F);
Inc(Index);
if P^ <> #0 then Inc(P, SepLen);
end;
if Index < ALen then SetLength(Result, Index);
end;
Sample usage:
var
res: TStringDynArray;
begin
res := Explode(':', yourString);

Answer 11

I wrote this function which returns linked list of separated strings by specific delimiter. Pure free pascal without modules.
Program split_f;
type
PTItem = ^TItem;
TItem = record
str : string;
next : PTItem;
end;
var
s : string;
strs : PTItem;
procedure split(str : string;delim : char;var list : PTItem);
var
i : integer;
buff : PTItem;
begin
new(list);
buff:= list;
buff^.str:='';
buff^.next:=nil;
for i:=1 to length(str) do begin
if (str[i] = delim) then begin
new(buff^.next);
buff:=buff^.next;
buff^.str := '';
buff^.next := nil;
end
else
buff^.str:= buff^.str+str[i];
end;
end;
procedure print(var list:PTItem);
var
buff : PTItem;
begin
buff := list;
while buff<>nil do begin
writeln(buff^.str);
buff:= buff^.next;
end;
end;
begin
s := 'Hi;how;are;you?';
split(s, ';', strs);
print(strs);
end.

Answer 12

Jedi Code Library provides an enhanced StringList with built-in Split function, that is capable of both adding and replacing the existing text. It also provides reference-counted interface. So this can be used even with older Delphi versions that have no SplitStrings and without careful and a bit tedious customizations of stock TStringList to only use specified delimiters.
For example given text file of lines like Dog 5 4 7 one can parse them using:
var slF, slR: IJclStringList; ai: TList<integer>; s: string; i: integer;
action: procedure(const Name: string; Const Data: array of integer);
slF := TJclStringList.Create; slF.LoadFromFile('some.txt');
slR := TJclStringList.Create;
for s in slF do begin
slR.Split(s, ' ', true);
ai := TList<Integer>.Create;
try
for i := 1 to slR.Count - 1 do
ai.Add(StrToInt(slR[i]));
action(slR[0], ai.ToArray);
finally ai.Free; end;
end;
http://wiki.delphi-jedi.org/wiki/JCL_Help:IJclStringList.Split#string#string#Boolean

Answer 13

This will solve your problem
interface
TArrayStr = Array Of string;
implementation
function SplitString(Text: String): TArrayStr;
var
intIdx: Integer;
intIdxOutput: Integer;
const
Delimiter = ';';
begin
intIdxOutput := 0;
SetLength(Result, 1);
Result[0] := '';
for intIdx := 1 to Length(Text) do
begin
if Text[intIdx] = Delimiter then
begin
intIdxOutput := intIdxOutput + 1;
SetLength(Result, Length(Result) + 1);
end
else
Result[intIdxOutput] := Result[intIdxOutput] + Text[intIdx];
end;
end;

Answer 14

My favourite function for splitting:
procedure splitString(delim: char; s: string; ListOfStrings: TStrings);
var temp: string;
i: integer;
begin
ListOfStrings.Clear;
for i:=1 to length(s) do
begin
if s[i] = delim then
begin
ListOfStrings.add(temp);
temp := '';
end
else
begin
temp := temp + s[i];
if i=length(s) then
ListOfStrings.add(temp);
end;
end;
ListOfStrings.add(temp);
end;

Answer 15

*
//Basic functionality of a TStringList solves this:
uses Classes //TStringList
,types //TStringDynArray
,SysUtils //StringReplace()
;
....
//--------------------------------------------------------------------------
function _SplitString(const s:string; const delimiter:Char):TStringDynArray;
var sl:TStringList;
i:integer;
begin
sl:=TStringList.Create;
//separete delimited items by sLineBreak;TStringlist will do the job:
sl.Text:=StringReplace(s,delimiter,sLineBreak,[rfReplaceAll]);
//return the splitted string as an array:
setlength(Result,sl.count);
for i:=0 to sl.Count-1
do Result[i]:=sl[i];
sl.Free;
end;
//To split a FileName (last item will be the pure filename itselfs):
function _SplitPath(const fn:TFileName):TStringDynArray;
begin
result:=_SplitString(fn,'\');
end;
*

Answer 16

The base of NGLG answer https://stackoverflow.com/a/8811242/6619626 you can use the following function:
type
OurArrayStr=array of string;
function SplitString(DelimeterChars:char;Str:string):OurArrayStr;
var
seg: TStringList;
i:integer;
ret:OurArrayStr;
begin
seg := TStringList.Create;
ExtractStrings([DelimeterChars],[], PChar(Str), seg);
for i:=0 to seg.Count-1 do
begin
SetLength(ret,length(ret)+1);
ret[length(ret)-1]:=seg.Strings[i];
end;
SplitString:=ret;
seg.Free;
end;
It works in all Delphi versions.

Answer 17

For delphi 2010, you need to create your own split function.
function Split(const Texto, Delimitador: string): TStringArray;
var
i: integer;
Len: integer;
PosStart: integer;
PosDel: integer;
TempText:string;
begin
i := 0;
SetLength(Result, 1);
Len := Length(Delimitador);
PosStart := 1;
PosDel := Pos(Delimitador, Texto);
TempText:= Texto;
while PosDel > 0 do
begin
Result[i] := Copy(TempText, PosStart, PosDel - PosStart);
PosStart := PosDel + Len;
TempText:=Copy(TempText, PosStart, Length(TempText));
PosDel := Pos(Delimitador, TempText);
PosStart := 1;
inc(i);
SetLength(Result, i + 1);
end;
Result[i] := Copy(TempText, PosStart, Length(TempText));
end;
You can refer to it as such
type
TStringArray = array of string;
var Temp2:TStringArray;
Temp1="hello:world";
Temp2=Split(Temp1,':')

Answer 18

procedure SplitCSV(S:STRING;out SL:TStringList);
var c,commatext:string;
a,b,up:integer;
begin
c:=s.Replace(' ','<SPACE>'); //curate spaces
//first ocurrence of "
a:=pos('"',c);
b:=pos('"',c,a+1);
if (a>0) and (b>0) then
begin
commatext:=commatext+copy(c,0,a-1);
commatext:=commatext+copy(c,a,b-a+1).Replace(',','<COMMA>'); //curate commas
up:=b+1;
end
else
commatext:=c;
//while continue discovering "
while (a>0) and (b>0) do
begin
a:=Pos('"',c,b+1);
b:=pos('"',c,a+1);
if (a>0) and (b>0) then
begin
commatext:=commatext+copy(c,up,a-up);
commatext:=commatext+copy(c,a,b-a+1).Replace(',','<COMMA>'); //curate commas
up:=b+1;
end;
end;
//last piece of text end
if up<c.Length then
commatext:=commatext+copy(c,up,c.Length-up+1);
//split text using CommaText
sl.CommaText:=commatext;
sl.Text:=sl.Text.Replace('<COMMA>',','); //curate commas
sl.Text:=sl.Text.Replace('<SPACE>',' '); //curate spaces
end;

Answer 19

interface
uses
Classes;
type
TStringArray = array of string;
TUtilStr = class
class function Split(const AValue: string; const ADelimiter: Char = ';'; const AQuoteChar: Char = '"'): TStringArray; static;
end;
implementation
{ TUtilStr }
class function TUtilStr.Split(const AValue: string; const ADelimiter: Char; const AQuoteChar: Char): TStringArray;
var
LSplited: TStringList;
LText: string;
LIndex: Integer;
begin
LSplited := TStringList.Create;
try
LSplited.StrictDelimiter := True;
LSplited.Delimiter := ADelimiter;
LSplited.QuoteChar := AQuoteChar;
LSplited.DelimitedText := AValue;
SetLength(Result, LSplited.Count);
for LIndex := 0 to LSplited.Count - 1 do
begin
Result[LIndex] := LSplited[LIndex];
end;
finally
LSplited.Free;
end;
end;
end.

Answer 20

I initially praised the answer from #Frank as I needed something that works for Delphi 6 and it appeared to work. However, I have since found that that solution has a bug whereby it still splits on #13#10 regardless of delimiter. Works perfectly if you are not expecting lines in your source string.
I wrote a simple parser that only works for single character delimiters. Note: it puts the values into a TStrings, not into an array as the op requested, but can easily be modified to adapt to arrays.
Here is the procedure:
procedure SplitString(const ASource: string; const ADelimiter: Char; AValues: TStrings);
var
i, lastDelimPos: Integer;
begin
AValues.Clear;
lastDelimPos := 0;
for i := 1 to Length(ASource) do
if ASource[i] = ADelimiter then
begin
if lastDelimPos = 0 then
AValues.Add(CopyRange(ASource, 1, i - 1))
else
AValues.Add(CopyRange(ASource, lastDelimPos + 1, i - 1));
lastDelimPos := i;
end;
if lastDelimPos = 0 then
AValues.Add(ASource)
else
AValues.Add(CopyRange(ASource, lastDelimPos + 1, MaxInt));
end;
function CopyRange(const s: string; const AIndexFrom, AIndexTo: Integer): string;
begin
Result := Copy(s, AIndexFrom, AIndexTo - AIndexFrom + 1);
end;
Note: as per C#'s string.Split(), a blank input string will result in a single blank string in the TStrings. Similarly, just having a delimiter by itself as the input string would result in two blank strings in the TStrings.
Here is the rough test code I used to ensure it's solid:
procedure AddTest(const ATestLine: string; const AExpectedResult: array of string);
var
expectedResult: TStringList;
i: Integer;
begin
expectedResult := TStringList.Create;
for i := 0 to Length(AExpectedResult) - 1 do
expectedResult.Add(AExpectedResult[i]);
testStrings.AddObject(ATestLine, expectedResult);
end;
//====================
AddTest('test', ['test']);
AddTest('', ['']);
AddTest(',', ['', '']);
AddTest('line1' + #13#10 + ',line 2,line3, line 4', ['line1' + #13#10, 'line 2', 'line3', ' line 4']);
AddTest('line1' + #13#10 + 'd,line 2,line3, line 4', ['line1' + #13#10 + 'd', 'line 2', 'line3', ' line 4']);
AddTest('line1,line 2,line3, line 4', ['line1', 'line 2', 'line3', ' line 4']);
AddTest('test, ', ['test', ' ']);
AddTest('test,', ['test', '']);
AddTest('test1,test2 ', ['test1', 'test2 ']);
AddTest('test1,test2', ['test1', 'test2']);
AddTest('test1,test2, ', ['test1', 'test2', ' ']);
AddTest('test1,test2,', ['test1', 'test2', '']);
//====================
testFailed := False;
for i := 0 to testStrings.Count - 1 do
begin
SplitString2(testStrings[i], ',', f);
log('Test ID=%d', [i]);
log(' Test String="%s"', [testStrings[i]]);
log(' Item count=%d', [f.Count]);
testResult := TStringList(TestStrings.Objects[i]);
if testResult.Count <> f.Count then
begin
Log('!!');
Log('!! Count mismatch. Got=%d, Expected=%d', [f.Count, testResult.Count]);
Log('!!');
testFailed := True;
end;
for j := 0 to f.Count - 1 do
begin
log(' Item %d="%s" (len=%d)', [j, f[j], Length(f[j])]);
if testResult[j] <> f[j] then
begin
Log('!!');
Log('!! Text mismatch. Got="%s", Expected="%s"', [f[j], testResult[j]]);
Log('!!');
testFailed := True;
end;
end;
end;
Edit: code for the CopyRange() function was missing, added now. My bad.