Why does this integer not work in an array alteration?

[Afrooman108]

E1:
Writeln('Choose seat character'); 
Readln(CH);                                         {Here the first letter of the array is decided}
Writeln('Choose seat number');
Readln(IP);                                          {This is the array number}

CH[IP].Surname:=Surname;                 {It struggles here With the IP variable)
CH[IP].name:=name;
CH[IP].date:=(Day,Month,Year);
CH[IP].az:=1;
__________
type
rec=record
Surname,name:string;
Date:string;
az:integer;
end;
__________

Var
  Fname, txt:string;
  Userfile:text;
  Q:Char;
  A:array[1..14] of rec;
  B:Array[1..16] of rec ;
  C:array[1..17] of rec  ;
  D:array[1..19] of rec   ;
  E:array[1..20] of rec    ;
  F:array[1..20] of rec     ;
  G:array[1..19] of rec      ;
  H:array[1..19] of rec       ;
  I:array[1..19] of rec        ;
  J:array[1..19] of rec         ;
  K:array[1..15] of rec          ;
  CH:char;
  IP:integer;
  P:integer                   ;
  Sel:string;
  Surname, Name, Date:string;
  Day,Month,Year:Integer;

[User137]

You can't CH[], because type of CH is not an array, but just char. You can't really refer to different arrays that easily, actually i can't find much of ways at all. You can use pointers and use case for 1 way:

Code: [Select]

var pCH: ^rec;

case CH of
  'A': pCH:=@A[IP];
 ..
end;
pCH^.Surname:=Surname;
pCH^.name:=name;
pCH^.date:=(Day,Month,Year);
pCH^.az:=1;But it raises questions on why would you want to have so many arrays?

PS. There are many spaces before some ; line ends.

[jshand2010]

E1:

Writeln('Choose seat character'); 
Readln(CH);                                         {Here the first letter of the array is decided}
Writeln('Choose seat number');
Readln(IP);

SetLength(IP, sizeof(IP));                                          {This is the array number}

CH[IP].Surname:=Surname;                 {It struggles here With the IP variable)
CH[IP].name:=name;
CH[IP].date:=(Day,Month,Year);
CH[IP].az:=1;
__________
type
rec=record
Surname,name:string;
Date:string;
az:integer;
end;
__________

Var
  Fname, txt:string;
  Userfile:text;
  Q:Char;
  A:array[1..14] of rec;
  B:Array[1..16] of rec ;
  C:array[1..17] of rec  ;
  D:array[1..19] of rec   ;
  E:array[1..20] of rec    ;
  F:array[1..20] of rec     ;
  G:array[1..19] of rec      ;
  H:array[1..19] of rec       ;
  I:array[1..19] of rec        ;
  J:array[1..19] of rec         ;
  K:array[1..15] of rec          ;
  CH: array of char;
  IP: array of integer;
  P:integer                   ;
  Sel:string;
  Surname, Name, Date:string;
  Day,Month,Year:Integer;

[Afrooman108]

Ok, cheers. I will just use 11 if statements.

How can i use the STR() function to turn 3 integer variables into a date.

Day:int;
Month:int ;
Year:int;
Date:str;

Date:=STR(DAY, '/', Month, '/', Year);

[Blaazen]

You can do:

Code: [Select]

Date:=inttostr(DAY)+ '/'+inttostr(Month)+'/'+inttostr(Year);
But we usually use TDateTime for storing dates.

Code: [Select]

procedure TForm1.Button1Click(Sender: TObject);
var date: TDateTime;
    y, m, d: Word;
begin
  y:=2013;
  m:=2;
  d:=12;
  date:=EncodeDate(y, m, d);
  writeln('Date is: '+DateToStr(date));
end;On Linux, you have to add the unit "clocale" to uses section of *.lpr file to have correct formatting.

[Afrooman108]

Hi, this is great. I have opted for the INTTOSTR option but i get identifier not found.

I will use the other method now. But i get this often, I'm using CRT.

Cheers

[User137]

Most string functions require SysUtils added to uses list. But with a little search you would know which unit it comes from.