我需要针对可用格式验证JSON。 JSON示例如下:
{
"sno": "1",
"project_name": "Abcd",
"contributors": [
{
"contributor_id": "1",
"contributor_name": "Ron"
},
{
"contributor_id": "2",
"contributor_name": "Dan"
}
],
"office": [ "Vegas", "New York" ]
}
在上面的示例中,我需要如下验证JSON:
如何根据上述条件解析JSON并检查所有键是否都具有有效值?
答案 0 :(得分:1)
您需要这样的对象:
public class MyObject
{
public string sno { get; set; }
public string project_name { get; set; }
public List<Contrbutor> contrbutors { get; set; }
public List<string> office { get; set; }
}
public class Contrbutor
{
public string contributor_id { get; set; }
public string contributor_name { get; set; }
}
通过JsonConvert对其进行解析
try
{
MyObject desObject = JsonConvert.DeserializeObject<MyObject>(yourJsonStringHere);
}
catch(Exception ex)
{
//IF PARSE IS NOT SUCCESSFUL CATCH THE PARSE EX HERE
}
,如果解析成功,则验证“ desObject”值。
答案 1 :(得分:1)
您可以构建自定义函数来检查json中各个键的值的数据类型。
1)将您的json解析为JObject
2)将此JObject
映射到您的SampleClass
。
3)然后,使用JTokenType
可以验证各个键的特定值是否为string
或array
或object
类型。
public string ValidateJson(string json)
{
JObject jObject = JObject.Parse(json);
SampleClass model = jObject.ToObject<SampleClass>();
string response = string.Empty;
foreach (var i in model.data)
{
switch (i.Key)
{
case "sno":
if (i.Value.Type != JTokenType.String)
response = "SNo is not a string";
break;
case "project_name":
if (i.Value.Type != JTokenType.String)
response = "Project name is not a string";
break;
case "contributors":
if (i.Value.Type != JTokenType.Array)
response = "Contributors is not an array";
break;
case "office":
if (i.Value.Type != JTokenType.Array)
response = "Office is not an array";
break;
}
}
return response;
}
您的SampleClass
将
class SampleClass
{
[JsonExtensionData]
public Dictionary<string, JToken> data { get; set; }
}