java中如何优化jason参数类型不匹配JSONObject obj = JSONObject.fromObject(new Student())用JSONUtil.deserialize()就可以了,范例: import org.apache.struts2.js2023-02-22Python140
java中如何优化jason参数类型不匹配JSONObject obj = JSONObject.fromObject(new Student())用JSONUtil.deserialize()就可以了,范例: import org.apache.struts2.js2023-02-22Python130