合并具有相同属性值的json对象c#

Posted 2023-04-12

【中文标题】合并具有相同属性值的json对象c#

【英文标题】：Merge json objects that with the same value of property c#

【发布时间】：2021-03-22 11:38:02

【问题描述】：

如何在具有相同键和值的同一响应中组合 JSON 对象。就像我有两个具有相同语言的对象：Python 我想将它们组合起来并列出该语言下的剩余数据 Python 我不希望它被重复

[
  [
    
      "language": "Python",
      "id": 319029846,
      "full_Name": "beurtschipper/Depix",
      "name": "Depix"
    ,
    
      "language": "Python",
      "id": 319169382,
      "full_Name": "benwilber/boltstream",
      "name": "boltstream"
    ,
    
      "language": "Python",
      "id": 316899719,
      "full_Name": "r0ysue/r0capture",
      "name": "r0capture"
    
  ],
  [
    
      "language": "YARA",
      "id": 318029147,
      "full_Name": "fireeye/red_team_tool_countermeasures",
      "name": "red_team_tool_countermeasures"
    
  ],
  [
    
      "language": "TypeScript",
      "id": 313443335,
      "full_Name": "pmndrs/valtio",
      "name": "valtio"
    
  ]
]

我想要的形式是什么

[
  [
    
      "language": "Python",
      "id": [319029846, 319169382, 316899719],
      "full_Name": ["beurtschipper/Depix", "benwilber/boltstream", "r0ysue/r0capture"],
      "name": ["Depix", "boltstream", "r0capture"]
    
  ],
  [
    
      "language": "YARA",
      "id": 318029147,
      "full_Name": "fireeye/red_team_tool_countermeasures",
      "name": "red_team_tool_countermeasures"
    
  ],
  [
    
      "language": "TypeScript",
      "id": 313443335,
      "full_Name": "pmndrs/valtio",
      "name": "valtio"
    
  ]
]

这是我正在使用的代码

    public class Items
    
        [JsonPropertyName("language")]
        public string Language  get; set; 
        [JsonPropertyName("id")]
        public int Id  get; set; 

        [JsonPropertyName("name")]
        public  string Name  get; set; 

        [JsonPropertyName("full_name")]
        public string Full_Name  get; set; 
        public string total_count  get; set; 
    

    public class Root
    
        [JsonPropertyName("items")]
        public List<Items> Items  get; set; 
    


        Root jObj2 = JsonConvert.DeserializeObject<Root>(readerResult);
        var result = jObj2.Items.Select(x => new
        
            x.Language,
            x.Id,
            x.Full_Name,
            x.Name
        ).GroupBy(x => x.Language).ToArray();
        return new JsonResult(result);

【参考方案1】：

GroupBy 是一个很好的起点。获得组后，您需要将每个组的各个属性选择到一个新列表中：

var result = jObj2.Items
    .GroupBy(x => x.Language)
    .Select(group => new
    
        Language = group.Key,
        Ids = group.Select(x => x.Id).ToList(),
        FullNames = group.Select(x => x.Full_Name).ToList(),
        Names = group.Select(x => x.Name).ToList()
    )
    .ToArray();

【讨论】：

哦，先生，我为这很容易感到羞愧，除非我尝试将这些 ID、名称..etc 列在 Ids = group.Select(x => x.Id).ToList(), 列表中，但我遇到了它们不是列表，我即使我应该在 Select 之前使用 GroupBy，但现在我有点知道 GroupBy 是如何工作的。如果你不介意我会问另一个问题.. 要知道的东西是经验还是我学得不好？当然非常感谢您的帮助。

你使用 Linq 的次数越多，你就越容易找到这类任务