Question

我有一个像这样的对象Person：

public class Person
{
    public string Name { get; set; }
    public int Age { get; set; }
    public string Address { get; set; }
}

我尝试创建一个这样的列表：

 var personList = new List<object>();
 personList.Add(new Person { Address = "addr1", Age = 20, Name = "Person1" });
 personList.Add(new Person { Address = "addr2", Age = 22, Name = "Person2" });
 personList.Add(new Person { Address = "addr3", Age = 25, Name = "Person1" });

 var jsonString = JsonConvert.SerializeObject(personList);

这是jsonString

的结果

[{"Name":"Person1","Age":20,"Address":"addr1"},

{"Name":"Person2","Age":22,"Address":"addr2"},

{"Name":"Person1","Age":25,"Address":"addr3"}]

以下是我的预期结果，那我该怎么做呢？

{
    Person1:{"Name":"Person1","Age":20,"Address":"addr1"},

    Person2:{"Name":"Person2","Age":22,"Address":"addr2"},

    Person3:{"Name":"Person3","Age":25,"Address":"addr3"}
}

Answer 1

当我没错的时候你必须使用{{ csrf_field() }} <div class="form-group"> <label for="imageInput" class="control-label col-sm-3">Upload Image</label> <div class="col-sm-9"> <input type="file" name="file"> </div> </div> <div class="form-group"> <div class="col-md-6-offset-2"> <input type="submit" class="btn btn-primary" value="Save"> </div> </div> </form>。这应该按照这种方式序列化，你想要它。

如果您不想要冗余数据，可以选择将Dictionary<string, Person>添加到[JsonIgnore]。

修改：您可以使用自定义JsonConverter将列表直接序列化到字典：Newtonsoft.Json serialize collection (with indexer) as dictionary

Answer 2

public class Person
{
    public string Name { get; set; }
    public int Age { get; set; }
    public string Address { get; set; }
    // constructor
    public Person(string name, int age, string address){
        this.Name = name;
        this.Age = age;
        this.Address = address;
    }
 }

然后

 List<person>People = new List<person>();
 People.Add(new Person("name1", 1, "address1"));
 People.Add(new Person("name2", 2, "address2"));
 People.Add(new Person("name3", 3, "address3"));


string jsonString = JsonConvert.SerializeObject(People);

然后在客户端上，省略了关于将其发送到客户端的部分

var people = JSON.parse(jsonString)

如何以不同的名称将同一对象的列表解析为JSON

2 个答案: