How to remove leading zero
Wasis Sugiono
For example:
'000' --> '0'
'001' --> '1'
'010' --> '10'
'100' --> '100'
'127.000.000.001' --> '127.0.0.1'
Thanks in advance.
David
function RemoveLeadingZeros(const S: String): String;
var
sTmp: String;
i: Integer;
begin
for i := 1 to Length(S) do
begin
if (S[i] = '.') or (i = Length(S)) then
begin
if i < Length(S) then
Result := Result + FormatFloat('##0', StrToFloat(sTmp)) + '.'
else
Result := Result + FormatFloat('##0', StrToFloat(sTmp + S[i]));
sTmp := '';
end
else
sTmp := sTmp + S[i];
end;
end;
David.
Brendan Delumpa
function StripFrontChars(S : String; ch : Char) : String;
begin
while (S[1] = ch) and (Length(S) > 0) do
S := Copy(S, 2, Length(S) - 1);
Result := S;
end;
--
Brendan
===========================================
If you think you have all the answers,
you haven't thought of all the questions...
===========================================
Wasis Sugiono wrote in message <36E4CA66...@rad.net.id>...
Philippe Ranger
How can I remove leading zeros of a string?
For example:
'000' --> '0'
'001' --> '1'
'010' --> '10'
'100' --> '100'
'127.000.000.001' --> '127.0.0.1'
<<
Function unZero (s: string): string;
Var
j, j0: integer;
qDel0: boolean;
Begin
qDel0 := true;
j := 1;
while (j <= length(s)) do begin
case s[j] of
'1'..'9': qDel0 := false;
'0': if qDel0 then begin
j0 := j;
while (j < length(s)) and
(s[j+1] in ['0'..'9']) and
(s[j] = '0')
do inc(j);
delete(s, j0, j-j0);
end;
else qDel0 := true;
end;
inc(j);
end;
result := s;
End;
PhR
Eyal Post
s:=IntToStr(StrToInt(s));
for the first 4 examples.
For the 5th - use the answers above.
Eyal
Wasis Sugiono wrote:
>
> How can I remove leading zeros of a string?
>
> For example:
> '000' --> '0'
> '001' --> '1'
> '010' --> '10'
> '100' --> '100'
> '127.000.000.001' --> '127.0.0.1'
>
> Thanks in advance.
--
Eyal Post
mailto:ey...@ladpc.gov.il
mailto:eyal...@geocities.com
------------
Peter Below (TeamB)
> How can I remove leading zeros of a string?
>
> For example:
> '000' --> '0'
> '001' --> '1'
> '010' --> '10'
> '100' --> '100'
>
Function RemoveLeadingZeros( Const S: String ): String;
Begin
Result := S;
While (Length(Result) > 1) and (Result[1] = '0') Do
Delete(Result, 1, 1 );
End;
> '127.000.000.001' --> '127.0.0.1'
This is a completely different problem, you could write a function that
takes the string apart into the numeric substring, passes each
substring to RemoveLeadingZeros and assembles the resulting
substrings again.
Function RemoveExtraZerosFromIPAddress( S: String ):String;
Var
dotpos: Integer;
Begin
Result := EmptyStr;
Repeat
dotpos := Pos('.', S);
If dotpos = 0 Then
Result := Result + S
Else Begin
Result := Result + RemoveLeadingZeros( Copy(S, 1, dotpos-1))
+ '.';
Delete(S, 1, dotpos);
End;
Until dotpos = 0;
End;
Peter Below (TeamB) 10011...@compuserve.com)
No e-mail responses, please, unless explicitely requested!
Mike Orriss
> while (S[1] = ch) and (Length(S) > 0) do
>
Doesn't that give you a problem?
I would do it the other way round:
while ((Length(S) > 0) and (S[1] = ch) do
Mike Orriss (m...@3kcc.co.uk)
http://www.3kcc.co.uk/notetree.htm
Philippe Ranger
How can I remove leading zeros of a string?
<<
Yesterday's code, simplified —
Function unZero (s: string): string;
(*Returns s with zeros removed iff they both come at the beginning or after
a non-digit, and precede a digit (0 included).*)
Var
j: integer;
nDel: integer; (*number of chars to delete in current run of 0s
-1 means we cannot delete the first 0 we meet*)
Begin
nDel := -1;
for j := length(s) downto 1 do case s[j] of
'0': inc(nDel);
'1'..'9': nDel := 0;
else begin
if (nDel > 0) then delete(s, j+1, nDel);
nDel := -1;
end;
end;
if (nDel > 0) then delete(s, 1, nDel);
Ernie Deel
message ...
> Function RemoveExtraZerosFromIPAddress( S: String ):String;
> Var
> dotpos: Integer;
> Begin
> Result := EmptyStr;
> Repeat
> dotpos := Pos('.', S);
> If dotpos = 0 Then
> Result := Result + S
> Else Begin
> Result := Result + RemoveLeadingZeros( Copy(S, 1,
dotpos-1))
> + '.';
> Delete(S, 1, dotpos);
> End;
> Until dotpos = 0;
> End;
>
With a little help from HyperString.<g>
procedure CondenseIPaddress(var S:AnsiString);
var
I:Integer;
begin
I := ScanFF(S,'00',1);
while I>0 do begin
if (I=1) or (S[I-1] = '.') then Delete(S,I,1) else
Inc(I,2);
I := ScanFF(S,'00',I);
end;
end;
--
Ernie Deel, EFD Systems
-----------------------------------------------
Any sufficiently advanced technology
is indistinguishable from a rigged demo.
Leicester Brad Ford Jr.
>Wazis: >>
>How can I remove leading zeros of a string?
Just wanted to throw my "hat" in...
>Function unZero (s: string): string;
>(*Returns s with zeros removed iff they both come at the beginning or after
> a non-digit, and precede a digit (0 included).*)
>Var
> j: integer;
> nDel: integer; (*number of chars to delete in current run of 0s
> -1 means we cannot delete the first 0 we meet*)
>Begin
> nDel := -1;
> for j := length(s) downto 1 do case s[j] of
> '0': inc(nDel);
> '1'..'9': nDel := 0;
> else begin
> if (nDel > 0) then delete(s, j+1, nDel);
> nDel := -1;
> end;
> end;
> if (nDel > 0) then delete(s, 1, nDel);
> result := s;
>End;
>
Wanted to try something easier to read... VAL and STRTOINT does
basically what you want (i.e. remove "0"s).
// *Assumption* no non numeric characters besides the "."
Function unZero (s: string): string; // works in TP
var i, j, p: integer; t, Result: string;
begin
s := s + '.'; Result := '';
repeat
p := pos('.', s);
if p = 0 then BREAK;
val(copy(s, 1, pred(p)), i, j);
str(i, t);
Result := Result + t + '.';
s := copy(s, succ(p), length(s));
until false;
Result := copy(Result, 1, length(Result) - 1);
unZero := Result;
end;
// Don't have DELPHI running now, so can't check for errors...
// But I believe the principle to be sound.
Function unZero (s: string): string; // works in DELPHI
var i, p: integer; t: string;
begin
s := s + '.'; Result := '';
repeat
p := pos('.', s);
if p = 0 then BREAK;
t := copy(s, 1, pred(p));
Result := Result + IntToStr(StrToInt(t)) + '.';
s := copy(s, succ(p), length(s));
until false;
Result := copy(Result, 1, length(Result) - 1);
end;
Thanks,
--------------------------------------------
Remove the "X" in the email address to reply
Philippe Ranger
<<
Pretty neat!
PhR
Philippe Ranger
Result := Result + IntToStr(StrToInt(t)) + '.';
<<
Nice. It translates the implicit specs more directly than does my code.
PhR
Philippe Ranger
Thank you, how was your vacation (was it a vacation?)
<<
Yes, mostly for my girlfriend. Skiing vacation with mucho rain...
PhR
Wasis Sugiono
IP := MaskEdit1.Text
MaskEdit1.EditMask := '099\.099\.099\.099;1;_'
So there must be 3 dots and at least a numeric in each gap
Function TProperties.IsValidIP(IP : String) : Boolean;
Var
i, j : Integer;
Addr : Array[1..4] of Integer;
Begin
IP := Trim(IP) + '.';
{Remove leading zeros}
j := 0;
Repeat
Inc(j);
i := Pos('.', IP);
Try
Addr[j] := StrToInt(Copy(IP, 1, i-1));
Result := Addr[j] in [0..255]
Except
Result := False;
End;
Delete(IP, 1, i);
Until (j > 3) OR (not Result);
If Result Then
ServerIP := Format('%d.%d.%d.%d', [Addr[1], Addr[2], Addr[3],
Addr[4]]);
End;
Leicester Brad Ford Jr.
Thank you, how was your vacation (was it a vacation?)
Myself, I am taking my wife (and daughter) to Las Vegas for a couple
of days to relax. I don't gamble but its good to get away from the
phone once in a while.
See ya,
David Arnall
CondenseIPaddress(011.100.001.011)
gives 011.100.001.011
D3 and HyperString v4.0
David
Ernie Deel wrote in message <7c68d1$jc...@forums.borland.com>...
Rick Rogers (TeamB)
> mostly for my girlfriend. Skiing vacation with mucho rain...
...an opportunity to sip hot cocoa in front of a blazing fire.
--
Rick Rogers (TeamB) | Fenestra Technologies
http://www.fenestra.com/
Ernie Deel
<7c7g0u$k7...@forums.borland.com>...
>Sorry, but this doesn't seam to work for me.
>
Check your implementation. The input string parameter must be
declared as "var".
Also, as given it only eliminates duplicate leading zeros. To
eliminate *all* leading zeros is similarly easy.
procedure CondenseIPaddress(var S:AnsiString);
var
I:Integer;
begin
I := 1;
while ScanW(S,'0#',I)>0 do
if (I=1) or (S[I-1] = '.') then Delete(S,I,1)
else Inc(I,2);
end;
Given S := '011.100.001.011' this function returns
'11.100.1.11' on my machine.
Leicester Brad Ford Jr.
>On Wed, 10 Mar 1999 22:06:10 -0500, "Philippe Ranger" <.> wrote:
>
>> mostly for my girlfriend. Skiing vacation with mucho rain...
>
>...an opportunity to sip hot cocoa in front of a blazing fire.
Now that sounds like a vacation.