如何从这个长期JSON文件采取的价值和新的事务后会增加自己,我需要最新的交易

问题描述 投票:-2回答:1

如何从这种在C#长JSON文件中获取价值目前我使用的动态变量在C#中,但如果我需要最新的JSON的交易将无法正常工作,因为一个特定位置的动态变量保持值,但其本地处理新后会增加交易,我需要的目的地标签所以我应该做到这一点。

{  
  "result":"success",
  "count":5,
  "transactions":[  
    {  
      "hash":"866B841C8C3B2D840497679FAC45A7D4341E12056AAA53F67471CA1B955FB58E",
      "ledger_index":37860930,
      "date":"2018-04-10T12:35:32+00:00",
      "tx":{  
        "TransactionType":"Payment",
        "Flags":2147483648,
        "SourceTag":0,
        "Sequence":20,
        "DestinationTag":0,
        "LastLedgerSequence":37860958,
        "Amount":"25000000",
        "Fee":"12",
        "SigningPubKey":"02917EB4353C539A2B0766EFAB1723E5BB744516C673FFCF21457DAD44BCECFE25",
        "TxnSignature":"304402202FA9FF18A572ACA1735B54CF5A69E10398EAF7003F3573E402ABA09B404F765C0220773481B494A941CC0F5338836A6B71F5FA2499943C41D50205C8EF158651BD86",
        "Account":"rLcqTPYFSJBWGmu2ZXh8JTYGgdXRPjWTrx",
        "Destination":"rfginNsDts73fZtzp858dT6Y6U67Y1QB7GV"
      },
      "meta":{  
        "TransactionIndex":1,
        "AffectedNodes":[  
          {  
            "CreatedNode":{  
              "LedgerEntryType":"AccountRoot",
              "LedgerIndex":"1C75CD43D75614E00025102355ADCCCF6302FAD5B82DC85F5AAF5696E4E39EC4",
              "NewFields":{  
                "Sequence":1,
                "Balance":"25000000",
                "Account":"rfginNsDt73fZtzp858dT6Y6U6s7Y1QB7GV"
              }
            }
          },
          {  
            "ModifiedNode":{  
              "LedgerEntryType":"AccountRoot",
              "PreviousTxnLgrSeq":37757840,
              "PreviousTxnID":"C794D81D17DF1EA735200E50B9CEADDF19160841E1321F08FDB7C7C4D6BC41AF",
              "LedgerIndex":"BE486CB1FED1B1805E4C42E8E96C73CFDA4E520C66BACEBBE441C1F8212D980E",
              "PreviousFields":{  
                "Sequence":20,
                "Balance":"537421240"
              },
              "FinalFields":{  
                "Flags":0,
                "Sequence":21,
                "OwnerCount":0,
                "Balance":"512421228",
                "Account":"rLcqTPYFSJBWGmu2ZXh8JTYGgdXRPjWTrx"
              }
            }
          }
        ],
        "TransactionResult":"tesSUCCESS",
        "delivered_amount":"25000000"
      }
    },
    {  
      "hash":"86E5D976DDCBF97047C86996286751B23C4384A8B4F6A04E38A57F30F54749CC",
      "ledger_index":38906860,
      "date":"2018-05-25T10:04:22+00:00",
      "tx":{  
        "TransactionType":"Payment",
        "Flags":2147483648,
        "Sequence":28534,
        "DestinationTag":2150,
        "Amount":"25000000",
        "Fee":"45",
        "SigningPubKey":"",
        "Account":"rG2uw1W9FXoiY4QiNqej5WeRxnoBEw4JbL",
        "Destination":"rfginNsDt73fZtzp858dT6Y6U67Y1QB7GV",
        "Signers":[  
          {  
            "Signer":{  
              "SigningPubKey":"032DCABB7A17A96E826435B461E474CB33C86C98B2281D304367FCFC6DE463A215",
              "TxnSignature":"3045022100F9802AE2F7800D42233A5B63999C0BD7AE5B7D60D8244C949FF81EA39959B4BC02205E37DD331ACE2637F2BEE758B5AD642D259F4140E8DE53EF610B6F58745BCC05",
              "Account":"rDc4zYyxMbXfUXNKw6zfM88RBUd9y8uPYX"
            }
          },
          {  
            "Signer":{  
              "SigningPubKey":"03019752C3ABCB2BC89EC5FAEF55D72AE940820A30FBA079D2BB8CE8841174EF21",
              "TxnSignature":"30450221009C7F9519912D494C9F009130927E1248BE6C2F877B1A913B49B8BFD926442F6802205377D1D242D07033DC6FDA8C93BC65638D636A87539F32A0D269AA9B537FCB0D",
              "Account":"rKB5N7FxvcwHSYq4NCKjyMpjML1MznMrFc"
            }
          }
        ]
      },
      "meta":{  
        "TransactionIndex":17,
        "AffectedNodes":[  
          {  
            "ModifiedNode":
c# json json.net json-rpc
1个回答
0
投票

有可用的金块包叫做Json.Net / Newtonsoft,这个包有方法可以很容易地进行序列化和反序列化JSON作为类对象。

步骤1:安装该模块>安装-封装Newtonsoft.Json -Version 11.0.2

步骤2:下一步是创建类,模仿你的完全一样的JSON所有关键要有相同的名称,你可以使用一个叫做选择性粘贴它可在Visual Studio中的功能。

创建一个空的类并复制JSON到剪贴板,然后点击编辑>选择性粘贴> JSON作为类,这会为你创建一类具有相同的结构,你的JSON。

步骤3:使用下面的代码反序列化JSON来一个类对象,

JSONClass jsonObjectWithValue = JsonConvert.Deserialize<JSONClass>(jsonString)
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