[英]Deserialize and parse a JSON string containing multiple objects
我正在使用Newtonsoft.Json,想反序列化和解析一个JSON像这样,在(数组?)中有多个点
{
"dataFile": {
"point": [
{
"ID": 1,
"time": "17.55",
"m_position": {
"x": 43.311744689941409,
"y": 147.30763244628907,
"z": 25.503124237060548
}
},
{
"ID": 2,
"time": "17.55",
"m_position": {
"x": 43.311744689941409,
"y": 147.30763244628907,
"z": 25.503124237060548
}
},
{
"ID": 3,
"time": "17.55",
"m_position": {
"x": 43.311744689941409,
"y": 147.30763244628907,
"z": 25.503124237060548
}
},
{
"ID": 4,
"time": "17.55",
"m_position": {
"x": 43.311744689941409,
"y": 147.30763244628907,
"z": 25.503124237060548
}
},
{
"ID": 5,
"time": "17.55",
"m_position": {
"x": 43.311744689941409,
"y": 147.30763244628907,
"z": 25.503124237060548
}
},
{
"ID": 6,
"time": "17.55",
"m_position": {
"x": 43.311744689941409,
"y": 147.30763244628907,
"z": 25.503124237060548
}
},
{
"ID": 7,
"time": "17.55",
"m_position": {
"x": 43.311744689941409,
"y": 147.30763244628907,
"z": 25.503124237060548
}
}
]
}
}
我用它来反序列化(strJSON是我上面粘贴的json代码):
var jPosData = JsonConvert.DeserializeObject<jsonPosSample>(strJSON);
我的jsonPosSample类是通过复制json并将其粘贴为Paste Special> Paste JSON as Classes生成的
所以我的jsonPosSample类看起来像这样:
class jsonPosSample
{
public class Rootobject
{
public Datafile dataFile { get; set; }
}
public class Datafile
{
public Point[] point { get; set; }
}
public class Point
{
public int ID { get; set; }
public string time { get; set; }
public M_Position m_position { get; set; }
}
public class M_Position
{
public float x { get; set; }
public float y { get; set; }
public float z { get; set; }
}
}
编辑:回答评论
通过反序列化来运行json代码:
var jPosData = JsonConvert.DeserializeObject<jsonPosSample>(strJSON);
debugOutput("Here's our JSON object: " + jPosData.ToString());
debugOutput打印出:
Here's our JSON object: Robbat.jsonPosSample
我正在努力了解如何将点存储在数组中或将其打印出来。 我需要制作6个对象,然后以某种方式将它们存储到数组中,对吗? 我该怎么做呢?
您提供了两个问题样本。
您的json末尾缺少'}'。
JsonConvert.DeserializeObject(strJSON); //由于您的datafile属性位于此类内。
如果要对jsonPosSample类进行操作,则
public class jsonPosSample
{
public Datafile dataFile { get; set; }
}
var result = JsonConvert.DeserializeObject<jsonPosSample>(strJSON);
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