早就可以了： User user = FastJson.getJson().parse(paraJson, User.class);

@JFinal 多谢波总，我试下

@JFinal 我这还是不能转，转换出的结果是：{} 测试样例：String pojson = FastJson.getJson().toJson(dicBusiness);//dicBusiness对象转换为json串：{"id":30,"name":"男","pid":0,"type":"XB"}
DicBusiness dic = FastJson.getJson().parse(pojson, DicBusiness.class);//再把json串（{"id":30,"name":"男","pid":0,"type":"XB"}）转换为DicBusiness对象，对象debug为{}；

样例的model对象：
public class DicBusiness extends BaseDicBusiness {
public static final DicBusiness dao = new DicBusiness().dao();
}
以及对象继承的BaseDicBusiness代码：
public abstract class BaseDicBusiness> extends Model implements IBean {

public M setId(java.lang.Long id) {
set("id", id);
return (M)this;
}

public java.lang.Long getId() {
return getLong("id");
}

public M setPid(java.lang.Long pid) {
set("pid", pid);
return (M)this;
}

public java.lang.Long getPid() {
return getLong("pid");
}

public M setType(java.lang.String type) {
set("type", type);
return (M)this;
}

public java.lang.String getType() {
return getStr("type");
}

public M setName(java.lang.String name) {
set("name", name);
return (M)this;
}

public java.lang.String getName() {
return getStr("name");
}

public M setSort(java.lang.Integer sort) {
set("sort", sort);
return (M)this;
}

public java.lang.Integer getSort() {
return getInt("sort");
}

public M setRemark(java.lang.String remark) {
set("remark", remark);
return (M)this;
}

public java.lang.String getRemark() {
return getStr("remark");
}

}

@JFinal json转换成DicBusiness对象为空

Jfinal3.3，我测试了，直接转model肯定是不行的。我的处理办法是，
String jsonString = HttpKit.readData(getRequest());
JSONObject dataMap = JSONObject.parseObject(jsonString);
Student model = new Student ();
model._setAttrs(dataMap.getJSONObject("student").toJavaObject(Map.class));
不知道还有没有更好的方式呢？