在 Delphi 中如何将动态数组保存到 FileStream？

Question

我在 Delphi 2009 中有以下结构：

type
  IndiReportIndi = record
    IndiName: string;
    NameNum: integer;
    ReportIndiName: string;
  end;

var
  XRefList: array of IndiReportIndi;

其中 XRefList 是动态数组。

我想将 XRefList 保存到 FileStream。我该如何做到这一点并包含所有 IndiName 和 ReportIndiName 字符串，以便稍后从该 FileStream 加载时可以再次检索它们？

Answer 1

type
  IndiReportIndi = record
    IndiName: string;
    NameNum: integer;
    ReportIndiName: string;
    procedure SaveToStream(Stream: TStream);
    procedure LoadFromStream(Stream: TStream);
  end;

type
  TXRefList = array of IndiReportIndi;

function LoadString(Stream: TStream): string;
var
  N: Integer;

begin
  Result:= '';
  Stream.ReadBuffer(N, SizeOf(Integer));
  if N > 0 then begin
    SetLength(Result, N);
//    Stream.ReadBuffer(Result[1], N * SizeOf(Char));
// fast version - see comment by A.Bouchez
    Stream.ReadBuffer(Pointer(Result)^, N * SizeOf(Char));
  end;
end;

procedure SaveString(Stream: TStream; const S: string);
var
  N: Integer;

begin
  N:= Length(S);
  Stream.WriteBuffer(N, SizeOf(Integer));
  if N > 0 then
//    Stream.WriteBuffer(S[1], N * SizeOf(Char));
// fast version - see comment by A.Bouchez
    Stream.WriteBuffer(Pointer(S)^, N * SizeOf(Char));
end;

procedure IndiReportIndi.LoadFromStream(Stream: TStream);
var
  S: string;

begin
  IndiName:= LoadString(Stream);
  Stream.ReadBuffer(NameNum, SizeOf(Integer));
  ReportIndiName:= LoadString(Stream);
end;

procedure IndiReportIndi.SaveToStream(Stream: TStream);
begin
  SaveString(Stream, IndiName);
  Stream.WriteBuffer(NameNum, SizeOf(Integer));
  SaveString(Stream, ReportIndiName);
end;

function LoadXRefList(Stream: TStream): TXRefList;
var
  N: Integer;
  I: Integer;

begin
  Stream.ReadBuffer(N, SizeOf(Integer));
  if N <= 0 then Result:= nil
  else begin
    SetLength(Result, N);
    for I:= 0 to N - 1 do
      Result[I].LoadFromStream(Stream);
  end;
end;

procedure SaveXRefList(Stream: TStream; const List: TXRefList);
var
  N: Integer;
  I: Integer;

begin
  N:= Length(List);
  Stream.WriteBuffer(N, SizeOf(Integer));
  for I:= 0 to N - 1 do
    List[I].SaveToStream(Stream);
end;

Answer 2

使用：http://code.google.com/p/kblib/

type
  IndiReportIndi = record
    IndiName: string;
    NameNum: integer;
    ReportIndiName: string;
  end;

  TXRefList = array of IndiReportIndi;

var
  XRefList: TXRefList;

保存整个 XRefList流式传输使用：

TKBDynamic.WriteTo(lStream, XRefList, TypeInfo(TXRefList));

加载回来：

TKBDynamic.ReadFrom(lStream, XRefList, TypeInfo(TXRefList));

Answer 3

var
  S: TStream;
  W: TWriter;
  A: array of IndiReportIndi;
  E: IndiReportIndi;
  ...
begin
  S := nil;
  W := nil;
  try 
    S := TFileStream.Create(...);
    W := TWriter.Create(S);
    W.WriteInteger(Length(A));
    for E in A do
    begin
      W.WriteString(E.IndiName);
      W.WriteInteger(E.NameNum);
      W.WriteString(E.ReportIndiName);
    end;
  finally
    W.Free;
    S.Free
  end;
end;  

要读取这些数据，您可以使用 TReader 和 ReadInteger/ReadString。

Answer 4

简单的 TFileStreamEx 保存变量

http://forum.chertenok.ru/viewtopic.php?t= 4642

从2005年开始就和他们一起工作，没有任何问题......

Answer 5

只是为了准确的答案：

考虑看看 TDynArray 包装器，它能够将任何记录序列化为二进制，也可以与动态数组进行序列化。

有很多类似 TList 的方法，包括新方法（例如散列或二分搜索）。

针对速度和磁盘使用进行了非常优化，适用于 Delphi 5 至 XE2 - 和开源。

另请参阅如何在 TList 中存储动态数组？