11

我有一个表示 JSON 的字符串,我想使用 JSON.NET 重命名一些属性。我需要一个通用函数来用于任何 JSON。就像是:

public static void Rename(JContainer container, Dictiontionary<string, string> mapping)
{
  foreach (JToken el in container.Children())
  {
    JProperty p = el as JProperty;
    if(el != null && mapping.ContainsKey(p.Name))
    {
      // **RENAME THIS NODE!!**
    }

    // recursively rename nodes
    JContainer pcont = el as JContainer;
    if(pcont != null)
    {
      Rename(pcont, mapping);
    }
  }
}

怎么做??

4

2 回答 2

19

我建议使用重命名的属性重建您的 JSON。我认为您不应该担心速度处罚,因为这通常不是问题。这是你如何做到的。

public static JToken Rename(JToken json, Dictionary<string, string> map)
{
    return Rename(json, name => map.ContainsKey(name) ? map[name] : name);
}

public static JToken Rename(JToken json, Func<string, string> map)
{
    JProperty prop = json as JProperty;
    if (prop != null) 
    {
        return new JProperty(map(prop.Name), Rename(prop.Value, map));
    }

    JArray arr = json as JArray;
    if (arr != null)
    {
        var cont = arr.Select(el => Rename(el, map));
        return new JArray(cont);
    }

    JObject o = json as JObject;
    if (o != null)
    {
        var cont = o.Properties().Select(el => Rename(el, map));
        return new JObject(cont);
    }

    return json;
}

这是一个使用示例:

var s = @"{ ""A"": { ""B"": 1, ""Test"": ""123"", ""C"": { ""Test"": [ ""1"", ""2"", ""3"" ] } } }";
var json = JObject.Parse(s);

var renamed = Rename(json, name => name == "Test" ? "TestRenamed" : name);
renamed.ToString().Dump();  // LINQPad output

var dict = new Dictionary<string, string> { { "Test", "TestRenamed"} };
var renamedDict = Rename(json, dict);
renamedDict.ToString().Dump();  // LINQPad output
于 2012-07-27T04:28:23.963 回答
5

我们使用这种方法。您可以使用 JObject 的 SelectToken() 找到所需的属性。是的,它确实支持 JsonPath。

public static class NewtonsoftExtensions
{
    public static void Rename(this JToken token, string newName)
    {
        var parent = token.Parent;
        if (parent == null)
            throw new InvalidOperationException("The parent is missing.");
        var newToken = new JProperty(newName, token);
        parent.Replace(newToken);
    }
}
于 2015-06-16T03:49:03.327 回答