C# 在 JObject 层次结构中按名称搜索特定 JToken
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Searching for a specific JToken by name in a JObject hierarchy
提问by Dima Serdechnyi
I have some Json response from server, for example:
我有一些来自服务器的 Json 响应，例如：
{"routes" : [
{
"bounds" : {
"northeast" : {
"lat" : 50.4639653,
"lng" : 30.6325177
},
"southwest" : {
"lat" : 50.4599625,
"lng" : 30.6272425
}
},
"copyrights" : "Map data ?2013 Google",
"legs" : [
{
"distance" : {
"text" : "1.7 km",
"value" : 1729
},
"duration" : {
"text" : "4 mins",
"value" : 223
},
And I want to get the value of token 'text' from
我想从
"legs" : [
{
"distance" : {
"text" : "1.7 km",
"value" : 1729
},
which is string with value "1.7 km".
这是值为“1.7 km”的字符串。
Question:is there any build-in function in NewtonsoftJson lib which can be look like:
问题：NewtonsoftJson lib 中是否有任何内置函数，如下所示：
public string(or JToken) GetJtokenByName(JObject document, string jtokenName)
or do I need to implement some recursive method which will search JToken by name in all JTokens and JArrays in JObject?
或者我是否需要实现一些递归方法，该方法将在 JObject 中的所有 JToken 和 JArray 中按名称搜索 JToken？
采纳答案by Brian Rogers
If you are looking for a very specific token and know the path to it, you can navigate to it easily using the built-in SelectToken()method. For example:
如果您正在寻找一个非常具体的令牌并知道它的路径，您可以使用内置SelectToken()方法轻松导航到它。例如：
string distance = jObject.SelectToken("routes[0].legs[0].distance.text").ToString();
If you need to find all occurences of a token with a given name in your JSON, no matter where they occur, then yes you'd need a recursive method. Here is one that might do the trick:
如果您需要在 JSON 中查找具有给定名称的令牌的所有出现，无论它们出现在何处，那么是的，您需要一个递归方法。这是一个可以解决问题的方法：
public static class JsonExtensions
{
public static List<JToken> FindTokens(this JToken containerToken, string name)
{
List<JToken> matches = new List<JToken>();
FindTokens(containerToken, name, matches);
return matches;
}
private static void FindTokens(JToken containerToken, string name, List<JToken> matches)
{
if (containerToken.Type == JTokenType.Object)
{
foreach (JProperty child in containerToken.Children<JProperty>())
{
if (child.Name == name)
{
matches.Add(child.Value);
}
FindTokens(child.Value, name, matches);
}
}
else if (containerToken.Type == JTokenType.Array)
{
foreach (JToken child in containerToken.Children())
{
FindTokens(child, name, matches);
}
}
}
}
Here is a demo:
这是一个演示：
class Program
{
static void Main(string[] args)
{
string json = @"
{
""routes"": [
{
""bounds"": {
""northeast"": {
""lat"": 50.4639653,
""lng"": 30.6325177
},
""southwest"": {
""lat"": 50.4599625,
""lng"": 30.6272425
}
},
""legs"": [
{
""distance"": {
""text"": ""1.7 km"",
""value"": 1729
},
""duration"": {
""text"": ""4 mins"",
""value"": 223
}
},
{
""distance"": {
""text"": ""2.3 km"",
""value"": 2301
},
""duration"": {
""text"": ""5 mins"",
""value"": 305
}
}
]
}
]
}";
JObject jo = JObject.Parse(json);
foreach (JToken token in jo.FindTokens("text"))
{
Console.WriteLine(token.Path + ": " + token.ToString());
}
}
}
Here is the output:
这是输出：
routes[0].legs[0].distance.text: 1.7 km
routes[0].legs[0].duration.text: 4 mins
routes[0].legs[1].distance.text: 2.3 km
routes[0].legs[1].duration.text: 5 mins
回答by mhand
This is pretty simple using the json paths and the SelectTokensmethod on JToken. This method is pretty awesome and supports wilds cards such as the following:
使用 json 路径和 上的SelectTokens方法非常简单JToken。这个方法非常棒，支持如下的通配符：
jObject.SelectTokens("routes[*].legs[*].*.text")
jObject.SelectTokens("routes[*].legs[*].*.text")
Check out this sample code:
查看此示例代码：
private class Program
{
public static void Main(string[] args)
{
string json = GetJson();
JObject jObject = JObject.Parse(json);
foreach (JToken token in jObject.SelectTokens("routes[*].legs[*].*.text"))
{
Console.WriteLine(token.Path + ": " + token);
}
}
private static string GetJson()
{
return @" {
""routes"": [
{
""bounds"": {
""northeast"": {
""lat"": 50.4639653,
""lng"": 30.6325177
},
""southwest"": {
""lat"": 50.4599625,
""lng"": 30.6272425
}
},
""legs"": [
{
""distance"": {
""text"": ""1.7 km"",
""value"": 1729
},
""duration"": {
""text"": ""4 mins"",
""value"": 223
}
},
{
""distance"": {
""text"": ""2.3 km"",
""value"": 2301
},
""duration"": {
""text"": ""5 mins"",
""value"": 305
}
}
]
}]}";
}
}
And here's the output:
这是输出：
routes[0].legs[0].distance.text: 1.7 km
routes[0].legs[0].duration.text: 4 mins
routes[0].legs[1].distance.text: 2.3 km
routes[0].legs[1].duration.text: 5 mins
回答by andersh
In case you want all values of a property, regardless of where it occurs, here is an alternative to recursion as described by @brian-rogers, using SelectTokenas suggested by @mhand:
如果您想要一个属性的所有值，无论它出现在哪里，这里是@brian-rogers 描述的递归的替代方法，使用SelectToken@mhand 的建议：
To get all values of duration.text, you can use SelectTokenand Linq:
要获取duration.text 的所有值，可以使用SelectToken和Linq：
var list = jObject.SelectTokens("$..duration.text")
.Select(t => t.Value<string>())
.ToList();
More info: Querying JSON with SelectToken
