Delphi isNumber - delphi

Is there a method in Delphi to check if a string is a number without raising an exception?
its for int parsing.
and an exception will raise if one use the
try
StrToInt(s);
except
//exception handling
end;

function TryStrToInt(const S: string; out Value: Integer): Boolean;
TryStrToInt converts the string S, which represents an integer-type number in either decimal or hexadecimal notation, into a number, which is assigned to Value. If S does not represent a valid number, TryStrToInt returns false; otherwise TryStrToInt returns true.
To accept decimal but not hexadecimal values in the input string, you may use code like this:
function TryDecimalStrToInt( const S: string; out Value: Integer): Boolean;
begin
result := ( pos( '$', S ) = 0 ) and TryStrToInt( S, Value );
end;

var
s: String;
iValue, iCode: Integer;
...
val(s, iValue, iCode);
if iCode = 0 then
ShowMessage('s has a number')
else
ShowMessage('s has not a number');

Try this function StrToIntDef()
From help
Converts a string that represents an integer (decimal or hex notation) to a number with error default.
Pascal
function StrToIntDef(const S: string; Default: Integer): Integer;
Edit
Just now checked the source of TryStrToInt() function in Delphi 2007. If Delphi 7 dont have this function you can write like this. Its just a polished code to da-soft answer
function TryStrToInt(const S: string; out Value: Integer): Boolean;
var
E: Integer;
begin
Val(S, Value, E);
Result := E = 0;
end;

XE4 and newer:
for ch in s do
TCharacter.IsNumber(ch);
Don't forget:
uses System.Character

In delphi 7 you can use the Val procedure. From the help:
Unit: System
Delphi syntax: procedure Val(S; var V; var Code: Integer);
S is a string-type expression; it must be a sequence of characters that form a signed real number.
V is an integer-type or real-type variable. If V is an integer-type variable, S must form a whole number.
Code is a variable of type Integer.
If the string is invalid, the index of the offending character is stored in Code; otherwise, Code is set to zero. For a null-terminated string, the error position returned in Code is one larger than the actual zero-based index of the character in error.

use this function
function IsNumber(N : String) : Boolean;
var
I : Integer;
begin
Result := True;
if Trim(N) = '' then
Exit(False);
if (Length(Trim(N)) > 1) and (Trim(N)[1] = '0') then
Exit(False);
for I := 1 to Length(N) do
begin
if not (N[I] in ['0'..'9']) then
begin
Result := False;
Break;
end;
end;
end;

For older Delphi versions from delphi 5 help example:
uses Dialogs;
var
I, Code: Integer;
begin
{ Get text from TEdit control }
Val(Edit1.Text, I, Code);
{ Error during conversion to integer? }
if Code <> 0 then
MessageDlg('Error at position: ' + IntToStr(Code), mtWarning, [mbOk], 0);
else
Canvas.TextOut(10, 10, 'Value = ' + IntToStr(I));
end;

In some languages decimal separators are different (for example, '.' is used in English and ',' is used in Russian). For these cases to convert string to real number the following procedure is proposed:
function TryStrToFloatMultiLang(const S : String; out Value : Extended) : Boolean;
var
dc : char;
begin
Result := false;
dc := DecimalSeparator;
DecimalSeparator := '.';
try
Result := TryStrToFloat(S, Value);
except
DecimalSeparator := ',';
Result := TryStrToFloat(S, Value);
end;
DecimalSeparator := dc;
end;
Update
As #Pep mentioned TryStrToFloat catch exceptions, but it returns boolean value. So the correct code is:
function TryStrToFloatMultiLang(const S : String; out Value : Extended) : Boolean;
var
dc : char;
begin
Result := false;
dc := DecimalSeparator;
DecimalSeparator := '.';
Result := TryStrToFloat(S, Value);
if not Result then begin
DecimalSeparator := ',';
Result := TryStrToFloat(S, Value);
end;
DecimalSeparator := dc;
end;

When you using procedure
val(s, i, iCode);
and set value xd ....
val('xd', i, iCode)
as a result we obtain: 13

standard unit Variants
function VarIsNumeric(v:Variant):Boolean

Related

How to get the string representation of a ShortCut Key including the SHIFTSTATE?

In a Delphi 10.4.2 Win32 VCL Application, and based on the question + solution here which provides a way to get the string representation of a Shortcut Key (but presumably with no possibility to also pass a SHIFTSTATE for the Shortcut Key) I wrote this code:
function MyGetSpecialShortcutName(ShortCut: TShortCut): string;
// gets shortcut name for e.g. VK_NUMPAD0 where TMenuItem.Shortcut gets the wrong shortcut name
var
ScanCode: Integer;
KeyName: array[0..255] of Char;
begin
Result := '';
FillChar(KeyName, SizeOf(KeyName), 0);
ScanCode := Winapi.Windows.MapVirtualKey(LoByte(Word(ShortCut)), 0) shl 16;
if ScanCode <> 0 then
begin
if Winapi.Windows.GetKeyNameText(ScanCode, KeyName, Length(KeyName)) <> 0 then
Result := KeyName;
end;
end;
function GetSpecialShortcutNameWithShiftState(const AScanCode: Word; const AShiftState: System.Classes.TShiftState = []): string;
begin
Result := MyGetSpecialShortcutName(Vcl.Menus.ShortCut(AScanCode, AShiftState));
end;
Usage:
Result := GetSpecialShortcutNameWithShiftState(VK_A, [ssCTRL]);
However, the Result is "A" where the expected Result should be "CTRL+A".
How to get the string representation of a ShortCut Key including the SHIFTSTATE?
The OP wants the key names fully localised, but for completeness I first show that the VCL already has a function to obtain a partly unlocalised string, namely, ShortCutToText in the Menus unit:
ShortCutToText(ShortCut(Ord('A'), [ssShift, ssAlt]))
This returns Shift+Alt+A on all systems.
Now, using the Win32 function GetKeyNameText already mentioned in the Q, it is easy to obtain a fully localised shortcut string:
function GetKeyName(AKey: Integer): string;
var
name: array[0..128] of Char;
begin
FillChar(name, SizeOf(name), 0);
GetKeyNameText(MapVirtualKey(AKey, 0) shl 16, #name[0], Length(name));
Result := name;
end;
function ModifierVirtualKey(AModifier: Integer): Integer;
begin
case AModifier of
Ord(ssShift):
Result := VK_SHIFT;
Ord(ssCtrl):
Result := VK_CONTROL;
Ord(ssAlt):
Result := VK_MENU;
else
Result := 0;
end;
end;
function ShortcutToString(AKey: Integer; AShiftState: TShiftState = []): string;
begin
Result := '';
for var Modifier in AShiftState do
begin
var ModifierKey := ModifierVirtualKey(Ord(Modifier));
if ModifierKey <> 0 then
Result := Result + IfThen(not Result.IsEmpty, '+') + GetKeyName(ModifierKey);
end;
Result := Result + IfThen(not Result.IsEmpty, '+') + GetKeyName(AKey);
end;
(Here I use a IfThen overload from StrUtils.)
Now,
ShortcutToString(Ord('A'), [ssShift, ssAlt])
returns SKIFT+ALT+A on my Swedish system. SKIFT is, as you might already have guessed, the Swedish name for the SHIFT key.

Why does this test to check if there are any illegal characters in a string keep coming up as false even though it should be true?

var
btof: boolean;
const
allowed = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789';
begin
btof:= false;
for i := 1 to length(sfilename) do
begin
if (sfilename[i] in [allowed[1] .. allowed[length(allowed)]])
then
btof := true;
end;
end;
I've been staring at this for the past hour trying to see the problem, but I just can't find it...
btof is always false, no matter the input.
The set notation [allowed[1] .. allowed[length(allowed)] boils down to a set of char ['a'..'9']. As the literal '9' has a lesser value than the literal 'a', the resulting set is just empty.
That is not how you check if a character is in an string. Actually you check if the character is between 'a' and '9'. That's not really what you intent to do.
Here is an implementation that works:
// Check if a filename contain any illegal character
function CheckIllegal(const SFilename : String) : Boolean;
var
C : Char;
const
Allowed = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789';
begin
Result := FALSE;
for C in SFilename do begin
if Pos(C, Allowed) < 1 then begin
Result := TRUE;
break;
end;
end;
end;
By the way, your list of allowed characters in a filename is incomplete, at least if you think about Windows OS.
var
bAllowed: boolean;
sfilename: string;
i, x: Integer;
C: Integer;
const
allowed = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789';
begin
bAllowed := true;
sfilename := edtInput.Text;
i := Length(sfilename);
x := 1;
while ((x <= i) and (bAllowed)) do
begin
bAllowed := Pos(sfilename[x], allowed) > 0;
Inc(x);
end;
if bAllowed then
ShowMessage('The string is valid')
else
ShowMessage('The string isn''t valid');
This is a more optimized version of the code. I take it that the purpose of the code that you gave was to determine if the string is valid and return a bool. Thus it only has to check up and till the point that it finds an illegal char in the string.
Note that I have changed it that if the string is valid it returns a true and invalid a false.

how to convert hexa to dec using delphi and hex to octal?

function HexToDec(Str: string): Integer;
var
i, M: Integer;
begin
Result:=0;
M:=1;
Str:=AnsiUpperCase(Str);
for i:=Length(Str) downto 1 do
begin
case Str[i] of
'1'..'9': Result:=Result+(Ord(Str[i])-Ord('0'))*M;
'A'..'F': Result:=Result+(Ord(Str[i])-Ord('A')+10)*M;
end;
M:=M shl 4;
end;
end;
procedure TForm1.Button1Click(Sender: TObject);
begin
if Edit1.Text<>'' then
Label2.Caption:=IntToStr(HexToDec(Edit1.Text));
end;
How to using it without function, because i want to call the result again in other line, and how about hexa to octal ? am i must conver from hexa to dec and then dec to octal?
Delphi can do this already, so you don't need to write a function parsing the number. It is quite simple, actually:
function HexToDec(const Str: string): Integer;
begin
if (Str <> '') and ((Str[1] = '-') or (Str[1] = '+')) then
Result := StrToInt(Str[1] + '$' + Copy(Str, 2, MaxInt))
else
Result := StrToInt('$' + Str);
end;
Note that that also handles negative hex numbers, or numbers like +$1234.
How to using it without function, because i want to call the result again in other line ?
If you want to re-use the value, assign the result of HexToDec to a variable and use that in IntToStr.
FWIW, in your function, there is no need to call AnsiUpperCase, because all hex digits fall in the ASCII range anyway. A much simpler UpperCase should work too.
My first comment would be that you are not converting hex to decimal with your function (although you are converting to decimal as an intermediate) but rather hex to integer. IntToStr then converts integer to base 10, effectively. To generalise what you want then I would create two functions - strBaseToInt and IntToStrBase where Base is meant to imply e.g. 16 for hex, 10 for dec, 8 for octal, etc., and assuming the convention adopted by hex that A=10, and so on but to (possibly) Z = 35 making the maximum base possible 36.
I don't handle + or - but that could be added easily.
In the reverse funtion, again for simplicity of illustration I have ommitted supporting negative values.
Edit
Thanks to Rudy for this improvement
Edit 2 - Overflow test added, as per comments
function StrBaseToInt(const Str: string; const Base : integer): Integer;
var
i, iVal, iTest: Longword;
begin
if (Base > 36) or (Base < 2) then raise Exception.Create('Invalid Base');
Result:=0;
iTest := 0;
for i:=1 to Length(Str) do
begin
case Str[i] of
'0'..'9': iVal := (Ord(Str[i])-Ord('0'));
'A'..'Z': iVal := (Ord(Str[i])-Ord('A')+10);
'a'..'z': iVal := (Ord(Str[i])-Ord('a')+10);
else raise Exception.Create( 'Illegal character found');
end;
if iVal < Base then
begin
Result:=Result * Base + iVal;
if Result < iTest then // overflow test!
begin
raise Exception.Create( 'Overflow occurred');
end
else
begin
iTest := Result;
end;
end
else
begin
raise Exception.Create( 'Illegal character found');
end;
end;
end;
Then, for example your HexToOct function would look like this
function HexToOct( Value : string ) : string;
begin
Result := IntToStrBase( StrBaseToInt( Value, 16), 8 );
end;
Additional
A general function would be
function BaseToBase( const Value : string; const FromBase, ToBase : integer ) : string;
begin
Result := IntToStrBase( StrBaseToInt( Value, FromBase ),ToBase );
end;

Only allow certain characters in a string

I am trying to validate a string, where by it can contain all alphebetical and numerical characters, aswell as the underline ( _ ) symbol.
This is what I tried so far:
var
S: string;
const
Allowed = ['A'..'Z', 'a'..'z', '0'..'9', '_'];
begin
S := 'This_is_my_string_0123456789';
if Length(S) > 0 then
begin
if (Pos(Allowed, S) > 0 then
ShowMessage('Ok')
else
ShowMessage('string contains invalid symbols');
end;
end;
In Lazarus this errors with:
Error: Incompatible type for arg no. 1: Got "Set Of Char", expected
"Variant"
Clearly my use of Pos is all wrong and I am not sure if my approach is even the correct way of going about it or not?
Thanks.
You will have to check every single character of the string, if it's contained in Allowed
e.g.:
var
S: string;
const
Allowed = ['A' .. 'Z', 'a' .. 'z', '0' .. '9', '_'];
Function Valid: Boolean;
var
i: Integer;
begin
Result := Length(s) > 0;
i := 1;
while Result and (i <= Length(S)) do
begin
Result := Result AND (S[i] in Allowed);
inc(i);
end;
if Length(s) = 0 then Result := true;
end;
begin
S := 'This_is_my_string_0123456789';
if Valid then
ShowMessage('Ok')
else
ShowMessage('string contains invalid symbols');
end;
TYPE TCharSet = SET OF CHAR;
FUNCTION ValidString(CONST S : STRING ; CONST ValidChars : TCharSet) : BOOLEAN;
VAR
I : Cardinal;
BEGIN
Result:=FALSE;
FOR I:=1 TO LENGTH(S) DO IF NOT (S[I] IN ValidChars) THEN EXIT;
Result:=TRUE
END;
If you are using a Unicode version of Delphi (as you seem to be), beware that a SET OF CHAR cannot contain all valid characters in the Unicode character set. Then perhaps this function will be useful instead:
FUNCTION ValidString(CONST S,ValidChars : STRING) : BOOLEAN;
VAR
I : Cardinal;
BEGIN
Result:=FALSE;
FOR I:=1 TO LENGTH(S) DO IF POS(S[I],ValidChars)=0 THEN EXIT;
Result:=TRUE
END;
but then again, not all characters (actually Codepoints) in Unicode can be expressed by a single character, and some characters can be expressed in more than one way (both as a single character and as a multi-character).
But as long as you constrain yourself within these limitations, one of the above functions should be useful. You can even include both, if you add an "OVERLOAD;" directive to the end of each function declaration, as in:
FUNCTION ValidString(CONST S : STRING ; CONST ValidChars : TCharSet) : BOOLEAN; OVERLOAD;
FUNCTION ValidString(CONST S,ValidChars : STRING) : BOOLEAN; OVERLOAD;
Lazarus/Free Pascal doesn't overload pos for that but has "posset" variants in unit strutils for that;
http://www.freepascal.org/docs-html/rtl/strutils/posset.html
Regarding Andreas' (IMHO correct ) remark, you can use isemptystr for that. It was meant to check for strings that only contain whitespace, but it basically checks if a string only contains characters in a set.
http://www.freepascal.org/docs-html/rtl/strutils/isemptystr.html
You can use Regular Expressions:
uses System.RegularExpressions;
if not TRegEx.IsMatch(S, '^[_a-zA-Z0-9]+$') then
ShowMessage('string contains invalid symbols');

How to count number of occurrences of a certain char in string?

How can I count the number of occurrences of a certain character in a string in Delphi?
For instance, assume that I have the following string and would like to count the number of commas in it:
S := '1,2,3';
Then I would like to obtain 2 as the result.
You can use this simple function:
function OccurrencesOfChar(const S: string; const C: char): integer;
var
i: Integer;
begin
result := 0;
for i := 1 to Length(S) do
if S[i] = C then
inc(result);
end;
Even though an answer has already been accepted, I'm posting the more general function below because I find it so elegant. This solution is for counting the occurrences of a string rather than a character.
{ Returns a count of the number of occurences of SubText in Text }
function CountOccurences( const SubText: string;
const Text: string): Integer;
begin
Result := Pos(SubText, Text);
if Result > 0 then
Result := (Length(Text) - Length(StringReplace(Text, SubText, '', [rfReplaceAll]))) div Length(subtext);
end; { CountOccurences }
And for those who prefer the enumerator loop in modern Delphi versions (not any better than the accepted solution by Andreas, just an alternative solution):
function OccurrencesOfChar(const ContentString: string;
const CharToCount: char): integer;
var
C: Char;
begin
result := 0;
for C in ContentString do
if C = CharToCount then
Inc(result);
end;
This one can do the work for if you're not handling large text
...
uses RegularExpressions;
...
function CountChar(const s: string; const c: char): integer;
begin
Result:= TRegEx.Matches(s, c).Count
end;
You can use the benefit of StringReplace function as:
function OccurencesOfChar(ContentString:string; CharToCount:char):integer;
begin
Result:= Length(ContentString)-Length(StringReplace(ContentString, CharToCount,'', [rfReplaceAll, rfIgnoreCase]));
end;
Simple solution and good performance (I wrote for Delphi 7, but should work for other versions as well):
function CountOccurences(const ASubString: string; const AString: string): Integer;
var
iOffset: Integer;
iSubStrLen: Integer;
begin
Result := 0;
if (ASubString = '') or (AString = '') then
Exit;
iOffset := 1;
iSubStrLen := Length(ASubString);
while (True) do
begin
iOffset := PosEx(ASubString, AString, iOffset);
if (iOffset = 0) then
Break;
Inc(Result);
Inc(iOffset, iSubStrLen);
end;
end;
Ummm... Am I missing something? Why not just...
kSepChar:=',';//to count commas
bLen:=length(sLineToCheck);
bCount:=0;//The numer of kSepChars seen so far.
bPosn:=1;//First character in string is at position 1
for bPosn:=1 to bLen do begin
if sLineToCheck[bPosn]=kSepChar then inc(bCount);
end;//

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