我需要反序列化从grogle maps api返回的这个json:
{
"destination_addresses": [
"Via Medaglie D'Oro, 10, 47121 Forlì FC, Italia",
"Via Torino, 20123 Milano, Italia",
"Via Guglielmo Marconi, 71, 40121 Bologna, Italia",
"Via Irnerio, 40126 Bologna, Italia"
],
"origin_addresses": [
"Via Medaglie D'Oro, 10, 47121 Forlì FC, Italia",
"Via Torino, 20123 Milano, Italia",
"Via Guglielmo Marconi, 71, 40121 Bologna, Italia",
"Via Irnerio, 40126 Bologna, Italia"
],
"rows": [
{
"elements": [
{
"distance": {
"text": "1 m",
"value": 0
},
"duration": {
"text": "1 min",
"value": 0
},
"status": "OK"
},
{
"distance": {
"text": "286 km",
"value": 286281
},
"duration": {
"text": "2 ore 48 min",
"value": 10083
},
"status": "OK"
},
{
"distance": {
"text": "80,1 km",
"value": 80088
},
"duration": {
"text": "1 ora 3 min",
"value": 3789
},
"status": "OK"
},
{
"distance": {
"text": "77,6 km",
"value": 77594
},
"duration": {
"text": "57 min",
"value": 3422
},
"status": "OK"
}
]
},
{
"elements": [
{
"distance": {
"text": "288 km",
"value": 287811
},
"duration": {
"text": "2 ore 48 min",
"value": 10052
},
"status": "OK"
},
{
"distance": {
"text": "1 m",
"value": 0
},
"duration": {
"text": "1 min",
"value": 0
},
"status": "OK"
},
{
"distance": {
"text": "212 km",
"value": 212423
},
"duration": {
"text": "2 ore 8 min",
"value": 7664
},
"status": "OK"
},
{
"distance": {
"text": "218 km",
"value": 218219
},
"duration": {
"text": "2 ore 9 min",
"value": 7740
},
"status": "OK"
}
]
},
{
"elements": [
{
"distance": {
"text": "78,5 km",
"value": 78528
},
"duration": {
"text": "56 min",
"value": 3346
},
"status": "OK"
},
{
"distance": {
"text": "212 km",
"value": 212190
},
"duration": {
"text": "2 ore 5 min",
"value": 7519
},
"status": "OK"
},
{
"distance": {
"text": "1 m",
"value": 0
},
"duration": {
"text": "1 min",
"value": 0
},
"status": "OK"
},
{
"distance": {
"text": "2,0 km",
"value": 1979
},
"duration": {
"text": "5 min",
"value": 316
},
"status": "OK"
}
]
},
{
"elements": [
{
"distance": {
"text": "74,7 km",
"value": 74719
},
"duration": {
"text": "55 min",
"value": 3278
},
"status": "OK"
},
{
"distance": {
"text": "218 km",
"value": 217951
},
"duration": {
"text": "2 ore 9 min",
"value": 7712
},
"status": "OK"
},
{
"distance": {
"text": "3,8 km",
"value": 3782
},
"duration": {
"text": "11 min",
"value": 671
},
"status": "OK"
},
{
"distance": {
"text": "1 m",
"value": 0
},
"duration": {
"text": "1 min",
"value": 0
},
"status": "OK"
}
]
}
],
"status": "OK"
}
我需要创建一个距离矩阵,所以我只对“距离”内的“值”字段感兴趣。
我尝试过这种方法:
DataSet data = JsonConvert.DeserializeObject<DataSet>(jsonResponse);
DataTable dataTab = data.Tables["Elements"];
foreach (DataRow elements in dataTab.Rows)
{
Console.WriteLine(elements["distance"]);
//Do something else here
}
但是JSonConvert在完成反序列化对象后返回“在JSON字符串中找到的附加文本。”
答案 0 :(得分:14)
您应该反序列化为与您的数据匹配的类。您可以在http://json2csharp.com/生成这些课程。
// use like
var rootObj = JsonConvert.DeserializeObject<RootObject>(jsonResponse);
foreach (var row in rootObj.rows)
{
foreach (var element in row.elements)
{
Console.WriteLine(element.distance.text);
}
}
// you might want to change the property names to .Net conventions
// use [JsonProperty] to let the serializer know the JSON names where needed
public class Distance
{
public string text { get; set; }
public int value { get; set; }
}
public class Duration
{
public string text { get; set; }
public int value { get; set; }
}
public class Element
{
public Distance distance { get; set; }
public Duration duration { get; set; }
public string status { get; set; }
}
public class Row
{
public List<Element> elements { get; set; }
}
public class RootObject
{
public List<string> destination_addresses { get; set; }
public List<string> origin_addresses { get; set; }
public List<Row> rows { get; set; }
public string status { get; set; }
}
答案 1 :(得分:1)
我认为问题在于根据我在搜索Stack Overflow时发现的类似问题对'DataSet'进行了演示。
尝试这个,因为我相信它会起作用(你可能需要使用'rows'而不是'Elements',但我相信使用JObject的方法将解决'附加文本'的主要问题。
JObject json = JsonConvert.DeserializeObject<JObject>(jsonResponse);
foreach (Dictionary<string, object> item in data["Elements"])
{
foreach (string val in item.Values) {
Console.WriteLine(val);
}
}
答案 2 :(得分:1)
使用动态:
dynamic json = JsonConvert.DeserializeObject(jsonResponse);
var rowCount = json.rows.Count;
Func<dynamic, int> getElementCount = r => r.elements.Count;
var maxElements = Enumerable.Max(json.rows, getElementCount);
var matrix = new int?[rowCount, maxElements];
for(int i = 0; i < rowCount; i++)
{
var elements = json.rows[i].elements;
for(int j = 0; j < elements.Count; j++)
{
var element = elements[j];
matrix[i, j] = element.distance.value;
}
}