使用 Delphi 构建 JSON

Question

我正在尝试实现一个函数，该函数返回一个包含类元素的 json 对象。这是我的功能：

procedure ListToJson(AInputList: TList<TRating>;
  AResponse: TJSONObject);
var
  i: Integer;
  jsonPair: TJSONPair;
  jsonObject: TJSONObject;
  jsonArray: TJSONArray;
begin
  jsonArray := TJSONArray.Create();
  jsonObject := TJSONObject.Create();
  jsonPair := TJSONPair.Create('ratings', jsonArray);

  for i := 0 to AInputList.Count - 1 do
  begin
    jsonObject.AddPair(TJSONPair.Create('idrating', IntToStr(AInputList[i].IdRating)));
    jsonObject.AddPair(TJSONPair.Create('idmark', IntToStr(AInputList[i].IdMark)));
    jsonObject.AddPair(TJSONPair.Create('value', IntToStr(AInputList[i].Value)));
    jsonObject.AddPair(TJSONPair.Create('description', AInputList[i].Description));
    jsonObject.AddPair(TJSONPair.Create('timeposted', FormatDateTime('yyyy-mm-dd hh:mm:ss', AInputList[i].TimePosted)));

    jsonArray.AddElement(jsonObject);
  end;

  AResponse.AddPair(jsonPair);
 end;

当我使用包含两个元素的列表对其进行测试时，返回的字符串是：

{
  "ratings":[{
      "idrating":"1",
      "idmark":"0",
      "value":"0",
      "description":"",
      "timeposted":"2015-07-29 11:25:03",
      "idrating":"2",
      "idmark":"0",
      "value":"0",
      "description":"",
      "timeposted":"2015-07-29 11:25:24"
  },{
      "idrating":"1",
      "idmark":"0",
      "value":"0",
      "description":"",
      "timeposted":"2015-07-29 11:25:03",
      "idrating":"2",
      "idmark":"0",
      "value":"0",
      "description":"",
      "timeposted":"2015-07-29 11:25:24"
  }]
} 

我尝试在每次循环迭代后删除所有对：

procedure ListToJson(AInputList: TList<TRating>;
  AResponse: TJSONObject);
var
  i: Integer;
  jsonPair: TJSONPair;
  jsonObject: TJSONObject;
  jsonArray: TJSONArray;
begin
  jsonArray := TJSONArray.Create();
  jsonObject := TJSONObject.Create();
  jsonPair := TJSONPair.Create('ratings', jsonArray);

  for i := 0 to AInputList.Count - 1 do
  begin
    jsonObject.AddPair(TJSONPair.Create('idrating', IntToStr(AInputList[i].IdRating)));
    jsonObject.AddPair(TJSONPair.Create('idmark', IntToStr(AInputList[i].IdMark)));
    jsonObject.AddPair(TJSONPair.Create('value', IntToStr(AInputList[i].Value)));
    jsonObject.AddPair(TJSONPair.Create('description', AInputList[i].Description));
    jsonObject.AddPair(TJSONPair.Create('timeposted', FormatDateTime('yyyy-mm-dd hh:mm:ss', AInputList[i].TimePosted)));

    jsonArray.AddElement(jsonObject);
    jsonObject.RemovePair('idrating');
    jsonObject.RemovePair('idmark');
    jsonObject.RemovePair('value');
    jsonObject.RemovePair('description');
    jsonObject.RemovePair('timeposted');
  end;

  AResponse.AddPair(jsonPair);
 end;

输出是一个包含 n（列表元素计数）个空对象的数组：{"ratings":[{},{}]}

我正在尝试构建的 json 应该是这样的：

{
  "ratings":[{
    "idrating":"1",
    "idmark":"0",
    "value":"0",
    "description":"",
    "timeposted":"2015-07-29 11:25:03"
  },{
    "idrating":"2",
    "idmark":"0",
    "value":"0",
    "description":"",
    "timeposted":"2015-07-29 11:25:24"
  }]
}

Answer 1

您有一个 jsonObject，因此每次将它添加到数组时都会重复它最终的任何状态。

在循环中构造jsonObject，然后你的数组中会有不同的对象。

  for i := 0 to AInputList.Count - 1 do
  begin
    jsonObject := TJSONObject.Create();
    jsonObject.AddPair(TJSONPair.Create('idrating', IntToStr(AInputList[i].IdRating)));
    jsonObject.AddPair(TJSONPair.Create('idmark', IntToStr(AInputList[i].IdMark)));
    jsonObject.AddPair(TJSONPair.Create('value', IntToStr(AInputList[i].Value)));
    jsonObject.AddPair(TJSONPair.Create('description', AInputList[i].Description));
    jsonObject.AddPair(TJSONPair.Create('timeposted', FormatDateTime('yyyy-mm-dd hh:mm:ss', AInputList[i].TimePosted)));

    jsonArray.AddElement(jsonObject);
  end;

或者，您可以定义一个函数来从 TRating 创建一个 TJSONObject，并在您的循环中使用它

function TRatingToJSON( Rating: TRating ): TJSONObject;
begin
  Result := TJSONObject.Create();

  Result.AddPair(TJSONPair.Create('idrating', IntToStr(Rating.IdRating)));
  Result.AddPair(TJSONPair.Create('idmark', IntToStr(Rating.IdMark)));
  Result.AddPair(TJSONPair.Create('value', IntToStr(Rating.Value)));
  Result.AddPair(TJSONPair.Create('description', Rating.Description));
  Result.AddPair(TJSONPair.Create('timeposted', FormatDateTime('yyyy-mm-dd hh:mm:ss', Rating.TimePosted)));
end;

procedure ListToJson(AInputList: TList<TRating>;
  AResponse: TJSONObject);
var
  i: Integer;
  jsonArray: TJSONArray;
begin
  jsonArray := TJSONArray.Create();

  for i := 0 to AInputList.Count - 1 do
  begin
    jsonArray.AddElement(TRatingToJSON(AInputList[i]));
  end;

  AResponse.AddPair(TJSONPair.Create('ratings', jsonArray));
 end;

Answer 2

是我，还是那真的很丑（和codous，或者很多代码的用词是什么）。因为我真的（真的）讨厌看到各种类型的长长的重载列表，并且真的（真的）喜欢Variant 类型，所以我创建了jsonDoc，它看起来像这样：

var
  x:array of OleVariant;
  i:integer;
begin
  SetLength(x,AInputList.Count);
  for i:=0 to AInputListCount-1 do
    x[i]:=JSON(
      ['idrating',AInputList[i].IdRating
      ,'idmark',AInputList[i].IdMark
      ,'value',AInputList[i].Value
      ,'description',AInputList[i].Description
      ,'timeposted',FormatDateTime('yyyy-mm-dd hh:mm:ss', AInputList[i].TimePosted)//VarFromDateTime?
      ]);
  end;
  AResponse:=JSON(['ratings',VarArrayOf(x)]);

Answer 3

找到解决方法：

procedure ListToJson(AInputList: TList<TRating>;
  AResponse: TJSONObject);
var
  i: Integer;
  jsonPair: TJSONPair;
  jsonObject: TList<TJSONObject>;
  jsonArray: TJSONArray;
begin
  jsonArray := TJSONArray.Create();
  jsonObject := TList<TJSONObject>.Create;
  jsonPair := TJSONPair.Create('ratings', jsonArray);

  try
    for i := 0 to AInputList.Count - 1 do
    begin
      jsonObject.Add(TJSONObject.Create());

      jsonObject.Last.AddPair(TJSONPair.Create('idrating', IntToStr(AInputList[i].IdRating)));
      jsonObject.Last.AddPair(TJSONPair.Create('idmark', IntToStr(AInputList[i].IdMark)));
      jsonObject.Last.AddPair(TJSONPair.Create('value', IntToStr(AInputList[i].Value)));
      jsonObject.Last.AddPair(TJSONPair.Create('description', AInputList[i].Description));
      jsonObject.Last.AddPair(TJSONPair.Create('timeposted', FormatDateTime('yyyy-mm-dd hh:mm:ss', AInputList[i].TimePosted)));

      jsonArray.AddElement(jsonObject.Last);
    end;

  AResponse.AddPair(jsonPair);
finally
  jsonObject.Free;
end;

结束；