我有一个JSON块,如下所示:
[
{
"id": 1,
"name": "Section1",
"project_id": 100,
"configs": [
{
"id": 1000,
"name": "myItem1",
"group_id": 1
}
]
},
{
"id": 2,
"name": "Section2",
"project_id": 100,
"configs": [
{
"id": 1001,
"name": "myItem2",
"group_id": 2
},
{
"id": 1002,
"name": "myItem3",
"group_id": 2
},
{
"id": 1003,
"name": "myItem4",
"group_id": 2
}
]
},
{
"id": 3,
"name": "Section3",
"project_id": 100,
"configs": [
{
"id": 1004,
"name": "myItem5",
"group_id": 5
},
]
}
]
我把它作为JArray把它拉进了Memory。
我需要迭代这一点,以便我只从配置子数组中获取id列表。理想情况下,我最终会得到这样的结果:
1000, myItem1
1001, myItem2
1002, myItem3
1003, myItem4
1004, myItem5
我很难理解Newstonsoft称之为JObject与JArray的对象,或者如何访问每个数据结构的各个部分。我现在的情况如下:
foreach (JObject config in result["configs"])
{
string id = (string)config["id"];
string name = (string)config["name"];
string gid = (string)config["group_id"];
Console.WriteLine(name + " - " + id + " - " + gid);
}
这不起作用,但我希望它说明了我的最终目标。我一直无法拼凑如何实现这一目标。
答案 0 :(得分:7)
JObject
是一个对象(类似于一个类):
{
"a": 1,
"b": true
}
JArray
是一个JSON数组,包含多个JObject
个实体:
[
{
"a": 1,
"b": true
},
{
"a": 2,
"b": true
}
]
JSON文档的根可以是对象或数组。在你的情况下,它是一个数组。
以下代码和fiddle表明您的代码没问题,只要您将文档反序列化为数组即可。
string json = "[{\"id\":1,\"name\":\"Section1\",\"project_id\":100,\"configs\":[{\"id\":1000,\"name\":\"myItem1\",\"group_id\":1}]},{\"id\":2,\"name\":\"Section2\",\"project_id\":100,\"configs\":[{\"id\":1001,\"name\":\"myItem2\",\"group_id\":2},{\"id\":1002,\"name\":\"myItem3\",\"group_id\":2},{\"id\":1003,\"name\":\"myItem4\",\"group_id\":2}]},{\"id\":3,\"name\":\"Section3\",\"project_id\":100,\"configs\":[{\"id\":1004,\"name\":\"myItem5\",\"group_id\":5},]}]";
JArray obj = Newtonsoft.Json.JsonConvert.DeserializeObject<JArray>(json);
foreach (var result in obj)
{
foreach (JObject config in result["configs"])
{
string id = (string)config["id"];
string name = (string)config["name"];
string gid = (string)config["group_id"];
Console.WriteLine(name + " - " + id + " - " + gid);
}
}