【发布时间】:2015-03-02 10:09:54
【问题描述】:
如何轻松地将 Java 对象转换/解析为 JSON 对象,即 com.couchbase.client.java.document.json.JsonObject 实例?
我尝试过这样做:
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;
}
并使用导入运行此行:
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);
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)
【问题讨论】: