Linq to JSON是用来操作JSON对象的.可以用于快速查询,修改和创建JSON对象.当JSON对象内容比较复杂,而我们仅仅需要其中的一小部分数据时,可以考虑使用Linq to JSON来读取和修改部分的数据而非反序列化全部.
在进行Linq to JSON之前,首先要了解一下用于操作Linq to JSON的类.
类名 | 说明 |
JObject |
用于操作JSON对象 |
JArray |
用语操作JSON数组 |
JValue |
表示数组中的值 |
JProperty |
表示对象中的属性,以"key/value"形式 |
JToken |
用于存放Linq to JSON查询后的结果 |
1.创建JSON对象
JObject staff = new JObject(); staff.Add(new JProperty("Name", "Jack")); staff.Add(new JProperty("Age", 33)); staff.Add(new JProperty("Department", "Personnel Department")); staff.Add(new JProperty("Leader", new JObject(new JProperty("Name", "Tom"), new JProperty("Age", 44), new JProperty("Department", "Personnel Department")))); Console.WriteLine(staff.ToString());
结果:
除此之外,还可以通过一下方式来获取JObject.JArray类似。
方法 | 说明 |
JObject.Parse(string json) |
json含有JSON对象的字符串,返回为JObject对象 |
JObject.FromObject(object o) |
o为要转化的对象,返回一个JObject对象 |
JObject.Load(JsonReader reader) |
reader包含着JSON对象的内容,返回一个JObject对象 |
2.创建JSON数组
JArray arr = new JArray(); arr.Add(new JValue(1)); arr.Add(new JValue(2)); arr.Add(new JValue(3)); Console.WriteLine(arr.ToString());
结果:
1.查询
首先准备Json字符串,是一个包含员工基本信息的Json
string json = "{\"Name\" : \"Jack\", \"Age\" : 34, \"Colleagues\" : [{\"Name\" : \"Tom\" , \"Age\":44},{\"Name\" : \"Abel\",\"Age\":29}] }";
①获取该员工的姓名
//将json转换为JObject JObject jObj = JObject.Parse(json); //通过属性名或者索引来访问,仅仅是自己的属性名,而不是所有的 JToken ageToken = jObj["Age"]; Console.WriteLine(ageToken.ToString());
结果:
②获取该员工同事的所有姓名
//将json转换为JObject JObject jObj = JObject.Parse(json); var names=from staff in jObj["Colleagues"].Children() select (string)staff["Name"]; foreach (var name in names) Console.WriteLine(name);
"Children()"可以返回所有数组中的对象
结果:
2.修改
①现在我们发现获取的json字符串中Jack的年龄应该为35
//将json转换为JObject JObject jObj = JObject.Parse(json); jObj["Age"] = 35; Console.WriteLine(jObj.ToString());
结果:
注意不要通过以下方式来修改:
JObject jObj = JObject.Parse(json); JToken age = jObj["Age"]; age = 35;
②现在我们发现Jack的同事Tom的年龄错了,应该为45
//将json转换为JObject JObject jObj = JObject.Parse(json); JToken colleagues = jObj["Colleagues"]; colleagues[0]["Age"] = 45; jObj["Colleagues"] = colleagues;//修改后,再赋给对象 Console.WriteLine(jObj.ToString());
结果:
3.删除
①现在我们想删除Jack的同事
JObject jObj = JObject.Parse(json); jObj.Remove("Colleagues");//跟的是属性名称 Console.WriteLine(jObj.ToString());
结果:
②现在我们发现Abel不是Jack的同事,要求从中删除
JObject jObj = JObject.Parse(json); jObj["Colleagues"][1].Remove(); Console.WriteLine(jObj.ToString());
结果:
4.添加
①我们发现Jack的信息中少了部门信息,要求我们必须添加在Age的后面
//将json转换为JObject JObject jObj = JObject.Parse(json); jObj["Age"].Parent.AddAfterSelf(new JProperty("Department", "Personnel Department")); Console.WriteLine(jObj.ToString());
结果:
②现在我们又发现,Jack公司来了一个新同事Linda
//将json转换为JObject JObject jObj = JObject.Parse(json); JObject linda = new JObject(new JProperty("Name", "Linda"), new JProperty("Age", "23")); jObj["Colleagues"].Last.AddAfterSelf(linda); Console.WriteLine(jObj.ToString());
结果:
使用函数SelectToken可以简化查询语句,具体:
①利用SelectToken来查询名称
JObject jObj = JObject.Parse(json); JToken name = jObj.SelectToken("Name"); Console.WriteLine(name.ToString());
结果:
②利用SelectToken来查询所有同事的名字
JObject jObj = JObject.Parse(json); var names = jObj.SelectToken("Colleagues").Select(p => p["Name"]).ToList(); foreach (var name in names) Console.WriteLine(name.ToString());
结果:
③查询最后一名同事的年龄
//将json转换为JObject JObject jObj = JObject.Parse(json); var age = jObj.SelectToken("Colleagues[1].Age"); Console.WriteLine(age.ToString());
结果:
FAQ
1.如果Json中的Key是变化的但是结构不变,如何获取所要的内容?
1 { 2 "trends": 3 { 4 "2013-05-31 14:31": 5 [ 6 {"name":"我不是谁的偶像", 7 "query":"我不是谁的偶像", 8 "amount":"65172", 9 "delta":"1596"}, 10 {"name":"世界无烟日","query":"世界无烟日","amount":"33548","delta":"1105"}, 11 {"name":"最萌身高差","query":"最萌身高差","amount":"32089","delta":"1069"}, 12 {"name":"中国合伙人","query":"中国合伙人","amount":"25634","delta":"2"}, 13 {"name":"exo回归","query":"exo回归","amount":"23275","delta":"321"}, 14 {"name":"新一吻定情","query":"新一吻定情","amount":"21506","delta":"283"}, 15 {"name":"进击的巨人","query":"进击的巨人","amount":"20358","delta":"46"}, 16 {"name":"谁的青春没缺失","query":"谁的青春没缺失","amount":"17441","delta":"581"}, 17 {"name":"我爱幸运七","query":"我爱幸运七","amount":"15051","delta":"255"}, 18 {"name":"母爱10平方","query":"母爱10平方","amount":"14027","delta":"453"} 19 ] 20 }, 21 "as_of":1369981898 22 }
其中的"2013-05-31 14:31"是变化的key,如何获取其中的"name","query","amount","delta"等信息呢?
通过Linq可以很简单地做到:
var jObj = JObject.Parse(jsonString); var tends = from c in jObj.First.First.First.First.Children() select JsonConvert.DeserializeObject(c.ToString()); public class Trend { public string Name { get; set; } public string Query { get; set; } public string Amount { get; set; } public string Delta { get; set; } }