json - 如何在Delphi Xe4中读取相当简单的JSON文件？

Question

我已经为此苦苦挣扎了一段时间，做一些简单的事情似乎花了太长时间。

我有一个这样的文件:

[
 {
  "FirstName": "Oleg",
  "Surname": "Buckley"
 },
 {
  "FirstName": "Amery",
  "Surname": "Mcmillan"
 },
 {
  "FirstName": "Denton",
  "Surname": "Burnett"
....

我希望能够将它们读入我的程序中。到目前为止，我已经完成了这个相当小的功能:

function GetGeneratedNames: TArray<string>;
var fileName: TFileName;
  JSONValue, jv: TJSONValue;
  JSONArray: TJSONArray;
  jo: TJSONObject;
  pair: TJSONPair;
begin
  result := nil;
  filename := ExePath + 'Names.json';
    JSONValue :=  TJSONObject.ParseJSONValue(TEncoding.ASCII.GetBytes(TFile.ReadAllText(filename)), 0);
    if JSONValue is TJSONArray then begin
     for jv in (JSONValue as TJSONArray) do begin
       if jv is TJSONObject then begin
         jo := jv as TJSONObject;
         for pair in jo do begin
           Append(result, jo.Value);
         end;
       end;
     end;
   end;
end{ GetGeneratedNames};

问题是，它返回一个空白字符串数组。谁能指出我正确的方向？

TIA标记

Answer 1

// XE5- version
uses System.SysUtils, Data.DBXJSON, System.IOUtils;

function GetGeneratedNames: TArray<string>;
var
  fileName: TFileName;
  JSONValue, jv: TJSONValue;
begin
  fileName := TPath.Combine(ExePath, 'Names.json');
  JSONValue := TJSONObject.ParseJSONValue(TFile.ReadAllText(fileName));
  try
    if JSONValue is TJSONArray then
    begin
      for jv in TJSONArray(JSONValue) do
      begin
        Append(Result, (jv as TJSONObject).Get('FirstName').JSONValue.Value);
        Append(Result, (jv as TJSONObject).Get('Surname').JSONValue.Value);
      end;
    end;
  finally
    JSONValue.Free;
  end;
end { GetGeneratedNames };

// XE6+ version
uses System.SysUtils, System.JSON, System.IOUtils;

function GetGeneratedNames: TArray<string>;
var
  fileName: TFileName;
  JSONValue, jv: TJSONValue;
begin
  fileName := TPath.Combine(ExePath, 'Names.json');
  JSONValue := TJSONObject.ParseJSONValue(TFile.ReadAllText(fileName));
  try
    if JSONValue is TJSONArray then
    begin
      for jv in TJSONArray(JSONValue) do
      begin
        Append(Result, jv.GetValue<string>('FirstName'));
        Append(Result, jv.GetValue<string>('Surname'));
      end;
    end;
  finally
    JSONValue.Free;
  end;
end { GetGeneratedNames };