如何轻松地将Java对象转换/解析为一个实例com.couchbase.client.java.document.json.JsonObject的JSON对象?
我尝试了以下代码:
import com.couchbase.client.deps.com.fasterxml.jackson.annotation.JsonProperty;
public class MyClass {
@JsonProperty("filed")
private String filed;
public MyClass(String filed) {
this.filed = filed;
}
public String getFiled() {
return filed;
}
并运行带导入的这些行:
import com.couchbase.client.deps.com.fasterxml.jackson.databind.ObjectMapper;
import com.couchbase.client.java.document.json.JsonObject;
ObjectMapper mapper = new ObjectMapper();
MyClass test = new MyClass("a");
JsonObject node = mapper.convertValue(test, JsonObject.class);
然后我得到:
java.lang.IllegalArgumentException: Unrecognized field "filed" (class com.couchbase.client.java.document.json.JsonObject), not marked as ignorable (one known property: "names"])
at [Source: N/A; line: -1, column: -1] (through reference chain: com.couchbase.client.java.document.json.JsonObject["filed"])
at com.couchbase.client.deps.com.fasterxml.jackson.databind.ObjectMapper._convert(ObjectMapper.java:2759)
at com.couchbase.client.deps.com.fasterxml.jackson.databind.ObjectMapper.convertValue(ObjectMapper.java:2685)