小编Gil*_*zov的帖子

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？

我试着这样做:

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)