这是一个基于How to iterate through a dictionary to get and pass key name to string的问题,下面的给定代码遍历JSON,获取键名和JArray索引并将它们有序地传递给JSON路径字符串,最后返回一个Dictionary(有序字符串, JsonValue),则字典的键名应按“ key1:key1-1:0”的顺序进行排序,这意味着requiredDictionary [“ key1:key1-1:0”] = commonDictionary [“ key1”] [“ key1-1 “] [0]。
根据以下JSON,如果 “五个”:{“ ArrayInFive”:[“ elem1”,“ elem2”]} 被删除,就可以正常工作。
C#代码
using Newtonsoft.Json;
using Newtonsoft.Json.Linq;
......
static void Main(string[] args)
{
var json = File.ReadAllText(@myJsonPath);
var jObj = JsonConvert.DeserializeObject<JObject>(json);
var desiredDict = FlattenJObjectToDictionary(jObj);
foreach (var key in desiredDict.Keys)
{
Console.WriteLine(key + " : " + desiredDict[key]);
}
Console.Read();
}
private static IDictionary<string, string> FlattenJObjectToDictionary(JObject obj)
{
// obtain a key/value enumerable and convert it to a dictionary
return NestedJObjectToFlatEnumerable(obj, null).ToDictionary(kv => kv.Key, kv => kv.Value);
}
private static IEnumerable<KeyValuePair<string, string>> NestedJObjectToFlatEnumerable(object data, string path = null)
{
JObject jObject = (JObject)data;
var jOP = jObject.Properties();
foreach (var jop in jOP)
{
if (jop.Value is JObject)
{
var child = (JObject)jop.Value;
// build the child path based on the root path and the property name
string childPath = path != null ? string.Format("{0}{1}:", path, jop.Name) : string.Format("{0}:", jop.Name);
// get each result from our recursive call and return it to the caller
foreach (var resultVal in NestedJObjectToFlatEnumerable(child, childPath))
{
yield return resultVal;
}
}
else if (jop.Value is JArray)
{
var jArray = (JArray)jop.Value;
for (int i = 0; i < jArray.Count; i++)
{
var child = jArray[i];
// build the child path based on the root path and the JArray index
string childPath = path != null ? string.Format("{0}{1}:{2}:", path, jop.Name, i.ToString()) : string.Format("{0}:{1}:", jop.Name, i.ToString());
// get each result from our recursive call and return it to the caller
foreach (var resultVal in NestedJObjectToFlatEnumerable(child, childPath))
{
yield return resultVal;
}
}
}
else
{
// this kind of assumes that all values will be convertible to string, so you might need to add handling for other value types
yield return new KeyValuePair<string, string>(string.Format("{0}{1}", path, Convert.ToString(jop.Name)), Convert.ToString(jop.Value));
}
}
}
JSON
{
"One": "Hey",
"Two": {
"Two": "HeyHey"
},
"Three": {
"Three": {
"Three": "HeyHeyHey"
}
},
"Four": [
{
"One": "Hey"
},
{
"Two":
{
"Two": "HeyHey"
}
}
],
"Five": {
"ArrayInFive": [ "elem1", "elem2" ]
}
}
我希望
desiredDictionary [“ Five”] [“ ArrayInFive”] [0] =“ elem1”
和
desiredDictionary [“ Five”] [“ ArrayInFive”] [1] =“ elem2”
但是弹出了“无法将JValue转换为JObject”的例外,我需要一些有关代码校正的帮助,也许是整个程序。
答案 0 :(得分:1)
将对JArray
中的NestedJObjectToFlatEnumerable
个对象的处理更改为:
else if (jop.Value is JArray)
{
var jArray = (JArray)jop.Value;
for (int i = 0; i < jArray.Count; i++)
{
var child = jArray[i];
if (child is JValue)
{
// return JValue objects directly as array elements instead of as objects in the array with their own property-value pairs
yield return new KeyValuePair<string, string>(string.Format("{0}{1}:{2}", path, jop.Name, i.ToString()), Convert.ToString(((JValue)child).Value));
}
else
{
// build the child path based on the root path and the JArray index
string childPath = path != null ? string.Format("{0}{1}:{2}:", path, jop.Name, i.ToString()) : string.Format("{0}:{1}:", jop.Name, i.ToString());
// get each result from our recursive call and return it to the caller
foreach (var resultVal in NestedJObjectToFlatEnumerable(child, childPath))
{
yield return resultVal;
}
}
}
}
这通过将元素作为数组的属性(具有由数组索引给定的属性名称)返回来处理数组元素为JValue
而不是具有其自身属性-值对的对象的情况到数组路径上。