Question

我以这种方式定义了namedQuery

@NamedQuery(name = "Emp.findAll", 
query = " select new test.entity.Emp(o.empNo, o.salary, p.projectId, p.projectName)
from Emp o inner join o.project p ")

在服务类中，我有

@GET
@Path("/projects")
@Produces(MediaType.APPLICATION_JSON)
public List<Emp> findAll() {
    return this.queryFindAll();

生成的JSON就是这种格式

[{"empNo":"2390","empName":"JAMES","projects":{"projectId":209,"projectName":"Z560"}}]

理想情况下，我想以下列格式生成JSON

[{"empNo":"2390","empName":"JAMES","projectId":"209","projectName":"Z560"}]

我该怎么做？

更新1

@GET
    @Path("/projects")
    @Produces(MediaType.APPLICATION_JSON)
    public Object findAll() throws IOException{
        String empAsJson = null;
        List<Emp> list = queryFindAll();
        ObjectMapper objectMapper = new ObjectMapper();
        empAsJson = objectMapper.writeValueAsString(list);        
        return empAsJson;
    }

Answer 1

看起来项目被序列化为地图/对象，你只需要将它作为地图或对象的数组。

使用伪代码更新：

List<Map<String,String>> rolledupEmps = new ArrayList<Map<String,String>>();
for(Emp emp : emps){
     Map<String,String> em = mapper.readValue(emp, HashMap.class);
     em.put("projectId",em.get("projects")?!=null:em.get("projects").get("projectId"),"");
     em.put("projectName",em.get("projects")?!=null:em.get("projects").get("projectName"),"");
     em.remove(""projects"");
     rolledupEmps.put(em);
}
return mapper.writeValueAsString(rolledupEmps);

生成的JSON格式错误

1 个答案: