3

我需要加载一个 JSON 文件，更改一个值，然后将其写回磁盘。

使用 SuperObject 很容易，但是如何使用 System.JSON 单元做同样的事情呢？

const
  PathToX = 'AllCategories.Category[0].subCategory[0].products[0].views.view[0].x';

var
  JsonFilename: string;

  JO: ISuperObject; // from the SuperObject unit
  JV: TJsonValue;   // from the System.Json unit

begin
  JsonFilename := ExtractFilePath(Application.ExeName)+'product.json');

  // Using the SuperObject unit:
  JO := SO(TFile.ReadAllText(JsonFilename));

  WriteLn('The old value of "x" is ', JO[PathToX].AsString);
  WriteLn('Changing value of x to "123"');
  JO.S[PathToX] := '123';   // Set the value of "x"
  WriteLn('The new value of "x" is ', JO[PathToX].AsString);

  // Now trying to do the same thing using the System.Json unit:
  JV := TJSONObject.ParseJsonValue(TFile.ReadAllText(JsonFilename));

  WriteLn('The old value of "x" is ', JV.GetValue<string>(PathToX));
  WriteLn('Changing value of x to "123"');
// Question: What code should go here to set the value of "x" using System.JSON ??? 
  WriteLn('The new value of "x" is ', JV.GetValue<string>(PathToX));

System.JSON 中似乎没有与“GetValue”方法等效的“SetValue”。

4

1 回答 1

7

TJSONObject确实支持类似于 SuperObject 的路径评估器。因此，您不必一次手动钻取一个对象的 JSON 值树（尽管如果您愿意，当然可以）。

然而，这些System.JSON类实际上并不是为修改现有数据而设计的（信不信由你）！它们专为解析数据和创建新数据而设计。所有表示简单值（整数、布尔值、字符串）的 JSON 类都是只读的。幸运的是，TJSONPair该类允许替换值，因此您必须利用它。

尝试这样的事情：

uses
  ..., System.JSON;

var
  JsonFilename: string;
  JV: TJSONValue;
  JO: TJSONObject;
  JoX: Integer;
  JoPair: TJSONPair;
begin
  JsonFilename := ExtractFilePath(Application.ExeName) + 'product.json';

  JV := TJSONObject.ParseJSONValue(TFile.ReadAllText(JsonFilename));
  if JV = nil then raise Exception.Create('Cannot parse file: ' + JsonFilename);
  try
    JO := JV as TJSONObject;

    JoX := JO.GetValue<Integer>('AllCategories.Category[0].subCategory[0].products[0].colors.color[0].views.view[0].x');
    WriteLn('The old value of "x" is ', JoX);

    WriteLn('Changing value of "x" to "123"');
    JoPair := JO.GetValue<TJSONObject>('AllCategories.Category[0].subCategory[0].products[0].colors.color[0].views.view[0]').Get('x');
    JoPair.JsonValue.Free;
    JoPair.JsonValue := TJSONNumber.Create(123);
    WriteLn('The new value of "x" is ', JoPair.JsonValue.Value);

    SaveAsDialog.FileName := JsonFilename;
    if SaveAsDialog.Execute then TFile.WriteAllText(SaveAsDialog.FileName, JO.ToJSON);
  finally
    JV.Free;
  end;
end;

或者：

uses
  ..., System.JSON;

var
  JsonFilename: string;
  JV: TJSONValue;
  JO: TJSONObject;
  JoX: TJSONPair;
begin
  JsonFilename := ExtractFilePath(Application.ExeName) + 'product.json';

  JV := TJSONObject.ParseJSONValue(TFile.ReadAllText(JsonFilename));
  if JV = nil then raise Exception.Create('Cannot parse file: ' + JsonFilename);
  try
    JO := JV as TJSONObject;

    JoX := JO.GetValue<TJSONObject>('AllCategories.Category[0].subCategory[0].products[0].colors.color[0].views.view[0]').Get('x');
    WriteLn('The old value of "x" is ', JoX.JsonValue.Value);

    WriteLn('Changing value of "x" to "123"');
    JoX.JsonValue.Free;
    JoX.JsonValue := TJSONNumber.Create(123);
    WriteLn('The new value of "x" is ', JoX.JsonValue.Value);

    SaveAsDialog.FileName := JsonFilename;
    if SaveAsDialog.Execute then TFile.WriteAllText(SaveAsDialog.FileName, JO.ToJSON);
  finally
    JV.Free;
  end;
end;
于 2015-10-30T02:12:39.553 回答