我有以下内容:
{"documents":
[{"keyPhrases":
[
"search results","Azure Search","fast search indexing","sophisticated search capabilities","Build great search experiences","time-sensitive search scenarios","service availability","managed service","service updates","index corruption","near-instantaneous responses","multiple languages","integrated Microsoft natural language stack","multiple indexes","application changes","ranking models","great relevance","years of development","primary interaction pattern","storage","Bing","data volume","rich","suggestions","hassle of dealing","Reliable throughput","website","incremental cost","complexity","faceting","traffic","mobile apps","business goals","users","applications","user expectations","Office"
],
"id":"1"}],
"errors":[]
}
我需要在keyPhrases中提取项目,但绝对不知道如何去做。
我尝试了以下内容:
KeyPhraseResult keyPhraseResult = new KeyPhraseResult();
/// <summary>
/// Class to hold result of Key Phrases call
/// </summary>
public class KeyPhraseResult
{
public List<string> keyPhrases { get; set; }
}
keyPhraseResult = JsonConvert.DeserializeObject<KeyPhraseResult>(content);
content
包含上面的JSON字符串。
然而keyPhraseResult返回一个空值。
任何身体都可以帮助我朝正确的方向发展吗?
谢谢。
答案 0 :(得分:4)
public class Document
{
public List<string> keyPhrases { get; set; }
public string id { get; set; }
}
public class RootObject
{
public List<Document> documents { get; set; }
public List<object> errors { get; set; }
}
你应该有这样的结构:
var result = JsonConvert.DeserializeObject<RootObject>(content);