如何从JSON对象中获取具有近似键的数值

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英文:

How to get a value from JSON object with approximate key

问题

我想从rackD1到rackD1000中心为south的键值使用LINQ提取。是否可以像JsonContent["serviceMap"][rackD1*]这样提取键值?

英文:

I have below JSON file under service map, The Key values are not fixed

{
"ServiceMap": {
    "rackAC1": {
        "Env": "Public",
        "Center": "north"
    },
    "rackD1": {
        "Env": "Public",
        "Center": "south"
    },
    "rackD2": {
        "Env": "Public",
        "Center": "North-south"
    },
    "rackD3": {
        "Env": "Public",
        "Center": "south"
    },
    ...,
    "rackD1000": {
        "Env": "Public",
        "Center": "south"
    },
    "rackBO": {
        "Env": "Public",
        "Center": "East"
    },
    "rackB1": {
        "Env": "Public",
        "Center": "West"
    }
}

}
I want to extract the key values from rackD1 to rackD1000 whose center is south using linq.
is it possible to extract the key value like JsonContent["serviceMap"][rackD1*]

答案1

得分: 1

只需将其反序列化为正确的类结构并在结果上使用 LINQ。

public class Root
{
    // 使用字典表示动态属性名称
    public Dictionary<string, ServiceMap> ServiceMap { get; set; }
}

public class ServiceMap
{
    public string Env { get; set; } // 如果不需要可以排除
    public string Center { get; set; }
}

然后筛选结果:

```csharp
var root = ... ; // 反序列化的 root
var result = root.ServiceMap
   .Where(kvp => kvp.Value.Center == "south") // 或 kvp.Value.Center.Equals("south", StringComparison.InvariantCultureIgnoreCase) 以进行不区分大小写的比较
   .Select(kvp => kvp.Key) // 键,如 "rackD1",...
   .ToList();

<details>
<summary>英文:</summary>

Just deserialize it to proper class structure and use LINQ on the results.

    public class Root
    {
        // use dictionary to represent dynamic property names
        public Dictionary&lt;string, ServiceMap&gt; ServiceMap { get; set; } 
    }

    public class ServiceMap
    {
        public string Env { get; set; } // can be excluded if not needed
        public string Center { get; set; }
    }

And filter results:

```csharp
var root = ... ; // deserialized root
var result = root.ServiceMap
   .Where(kvp =&gt; kvp.Value.Center == &quot;south&quot;) // or kvp.Value.Center.Equals(&quot;south&quot;, StringComparison.InvariantCultureIgnoreCase) for case-insensitive
   .Select(kvp =&gt; kvp.Key) // keys like &quot;rackD1&quot;, ...
   .ToList();

答案2

得分: 1

你不需要任何类来获取键列表:

List<string> keys = ((JObject)JObject.Parse(json)["ServiceMap"]).Properties()
    .Where(p => p.Name.Contains("rackD") 
                  && ((string)p.Value["Center"]).ToLower() == "south")
    .Select(p => p.Name)
    .ToList();
英文:

you dont need any classes to get list of keys

	List&lt;string&gt; keys = ((JObject)JObject.Parse(json)[&quot;ServiceMap&quot;]).Properties()
	.Where(p =&gt; p.Name.Contains(&quot;rackD&quot;) 
                  &amp;&amp; ((string)p.Value[&quot;Center&quot;]).ToLower() == &quot;south&quot;)
	.Select(p =&gt; p.Name)
	.ToList();

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  • 本文由 发表于 2023年2月6日 20:26:47
  • 转载请务必保留本文链接:https://go.coder-hub.com/75361306.html
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