C# 使用Newtonsoft.JSON将两个数组转换为一个JSON对象

我有数组

name[]

和

lastname[]

。如何将它们组合并转换为JSON字符串？我希望它是以下格式。我需要JSON中的“雇员”标题

{ "Employees" : [
    {"name": "John", "lastname": "Coleman"},
    {"name": "Chip", "lastname": "Dale"},
    {"name": "Ann", "lastname": "Smith"},
    {"name": "Terry", "lastname": "Johnson"},
    {"name": "Mary", "lastname": "Loggins"},
    {"name": "Timothy", "lastname": "Lopez"},
    {"name": "Jessica", "lastname": "Brown"}
]}

我需要一种有效的方法来实现这一点，因为数组中有很多项。实际上，我需要将两个以上的数组组合成一个JSON对象。为了简单起见，我用两个演示了我想要什么。它们都有相同数量的物品，并且都是订购的。我不想迭代数组并自己构造JSON字符串

更新：

我忘了提到我的数组是

IEnumerable

字符串数组和整数数组。下面是我尝试的，数组是在另一个类中创建的

  public string[] Name {
                get{ return  (Employees ?? Enumerable.Empty<Employee> ()).Select (p => p.name).ToArray(); }
            }

    public string[] Lastname {
                get{ return  (Employees ?? Enumerable.Empty<Employee> ()).Select (p => p.lastname).ToArray(); }
            }

    public int[] Age {
                get{ return  (Employees ?? Enumerable.Empty<Employee> ()).Select (p => p.age).ToArray(); }
            }

出于某种原因，这会返回类似于

{"Employees":[{"name":["John","Chip","Ann","Terry"],"lastname":["Coleman","Dale","Smith","Johnson"],"age":[42, 26, 33, 24]}]}

所有的名字，姓氏都放在一起。如何获得正确的格式

您可以将它们与组合成，然后序列化：

        string[] name = new string[] { "John", "Chip" };
        string[] lastname = new string[] { "Coleman", "Dale" };

        var employees = new { Employees = name.Zip(lastname, (n1, n2) => new { name = n1, lastname = n2 }) };
        var json = JsonConvert.SerializeObject(employees, Formatting.Indented);
        Debug.WriteLine(json);

哪些产出：

对于多个数组，并行遍历数组可能更容易：

        string[] name = new string[] { "John", "Chip" };
        string[] lastname = new string[] { "Coleman", "Dale" };
        string[] title = new string[] { "Mr", "Dr" };
        string[] profession = new string[] { "Coder", "Doctor" };

        var employees2 = new { Employees = Enumerable.Range(0, name.Length).Select(i => new { title = title[i], name = name[i], lastname = lastname[i], profession = profession[i] }) };
        var json2 = JsonConvert.SerializeObject(employees2, Formatting.Indented);
        Debug.WriteLine(json2);

更新

如果字符串位于

IEnumerable

中，则可以将外部可枚举项转换为数组，然后索引到其中。例如，给定以下测试用例：

        string[] name = new string[] { "John", "Chip" };
        string[] lastname = new string[] { "Coleman", "Dale" };
        string[] title = new string[] { "Mr", "Dr" };
        string[] profession = new string[] { "Coder", "Doctor" };

        IEnumerable<string[]> strings = new[] { title, name, lastname, profession };

其结果是：

更新2

如果实际有一个包含员工枚举的对象枚举，则可以使用。例如，给定以下类：

public class Employee
{
    public string name { get; set; }
    public string lastname { get; set; }
    public int age { get; set; }
    public string someMoreDataThatShouldNotBeSerialized { get; set; }
}

public class EmployeeContainer
{
    public IEnumerable<Employee> Employees { get; set; }
}

---

正在工作。

请向我们展示您的

名称

和

姓氏

数组。更新示例中的

X

是什么？它是一个包含员工列表的类上的可枚举对象吗？它是一个不同的反序列化json对象吗？它是一个对象还是一个包含属性的对象的可枚举对象<代码>公共字符串[]名称和<代码>公共字符串[]Lastname和<代码>公共int[]年龄？如果是这样的话，您有字符串数组，而不是字符串数组。您需要使用，比如说，

SelectMany

，将它们展平。是的，它是一个包含属性的可枚举项。如果我使用公共字符串[]Name{get{return（Employees？？Enumerable.Empty（））.SelectMany（p=>p.Name）；}}，我会得到错误CS0411：无法从用法推断方法'System.Linq.Enumerable.SelectMany（this System.Collections.Generic.IEnumerable，System.Func'）的类型参数。尝试显式指定类型参数这是否适用于2个以上的数组？Zip（）只接受2个数组作为参数。对于7+阵列，您是否会在彼此之后运行多个拉链？有更优雅的解决方案吗？@DanielHoffmann Mitscherling-好吧，因为它们是数组，你可以做

Enumerable.Range（0，array0.Length）。选择（i=>new{a=array0[i]，b=array1[i]，c=array2[i]，…）

否则就是多个ZIP，或者用多个输入创建你自己的扩展方法。对，第一个很明显，但是作者要求不要迭代数组，但是Zip（）无论如何都会这样做（但您不必为此编写代码）。我不知道他/她到底想尝试什么。很高兴知道你可以串多个拉链！（尽管扩展方法更好）@DanielHoffmann Mitscherling-我认为这意味着OP不想编写手动

foreach

循环。在某种级别的代码（可能是Json.NET本身）中，需要访问每个数组元素。@dbc同意！对你答案的编辑使它成为一个很好的答案。

        string[] name = new string[] { "John", "Chip" };
        string[] lastname = new string[] { "Coleman", "Dale" };
        string[] title = new string[] { "Mr", "Dr" };
        string[] profession = new string[] { "Coder", "Doctor" };

        IEnumerable<string[]> strings = new[] { title, name, lastname, profession };

        var stringArray = strings.ToArray();

        var employees2 = new { Employees = Enumerable.Range(0, name.Length).Select(i => new { title = stringArray[0][i], name = stringArray[1][i], lastname = stringArray[2][i], profession = stringArray[3][i] }) };
        var json2 = JsonConvert.SerializeObject(employees2, Formatting.Indented);
        Debug.WriteLine(json2);

{
  "Employees": [
    {
      "title": "Mr",
      "name": "John",
      "lastname": "Coleman",
      "profession": "Coder"
    },
    {
      "title": "Dr",
      "name": "Chip",
      "lastname": "Dale",
      "profession": "Doctor"
    }
  ]
}

public class Employee
{
    public string name { get; set; }
    public string lastname { get; set; }
    public int age { get; set; }
    public string someMoreDataThatShouldNotBeSerialized { get; set; }
}

public class EmployeeContainer
{
    public IEnumerable<Employee> Employees { get; set; }
}

        var X = GetAllEmployees();

        var employees = X.SelectMany(s => s.Employees ?? Enumerable.Empty<Employee>()).Select(e => new { name = e.name, lastname = e.lastname, age = e.age });
        var json = JsonConvert.SerializeObject(employees, Formatting.Indented);
        Debug.WriteLine(json);

    public static IEnumerable<EmployeeContainer> GetAllEmployees()
    {
        return new[] { 
            new EmployeeContainer { 
                Employees = 
                    new[] { 
                        new Employee { name = "John", lastname = "Coleman", age = 42, someMoreDataThatShouldNotBeSerialized = "someMoreData1" },
                        new Employee { name = "Chip", lastname = "Dale", age = 26, someMoreDataThatShouldNotBeSerialized = "someMoreData2" },
                    } 
            },
            new EmployeeContainer { 
                Employees = 
                    new[] { 
                        new Employee { name = "Ann", lastname = "Smith", age = 33, someMoreDataThatShouldNotBeSerialized = "someMoreData3" },
                        new Employee { name = "Terry", lastname = "Johnson", age = 24, someMoreDataThatShouldNotBeSerialized = "someMoreData4" }, 
                    } 
            },
            new EmployeeContainer()
        };
    }

[
  {
    "name": "John",
    "lastname": "Coleman",
    "age": 42
  },
  {
    "name": "Chip",
    "lastname": "Dale",
    "age": 26
  },
  {
    "name": "Ann",
    "lastname": "Smith",
    "age": 33
  },
  {
    "name": "Terry",
    "lastname": "Johnson",
    "age": 24
  }
]