从@ResponseBody创建JsonObject，而不使用gson库

Question

@ResponseBody返回

[{"id":1010,"name":"projectname2"}] type json string

但是我需要一个json字符串

[{"id":1010,"name":"projectname2","age":"21"}]

那么我如何连接属性到默认生成的json字符串?

我使用java spring-mvc框架与spring-json jar

@RequestMapping(value = "/projectsByEmployeeId/list", method = RequestMethod.GET, produces = MediaType.APPLICATION_JSON_VALUE)
@ResponseBody
public List<Project> getProjectsByEmployeeId(
        HttpServletRequest request) {
    String filter = request.getParameter("filter");
    FilterAttributes[] filterAttributes = null;
    try {
        filterAttributes = new ObjectMapper().readValue(filter,
                FilterAttributes[].class);
    } catch (Exception exception) {
        logger.error("Filtering parameter JSON string passing failed",
                exception);
    }
    if ((filterAttributes != null)
            && (!filterAttributes[0].getStringValue().isEmpty())) {
        return utilityService.getProjectsByEmployeeId(44L);//this is an example id
    } else {
        return utilityService.getProjects();
    }
}

Answer 1

另一种完成此操作的方法是纯字符串操作-

String json = "[{"id":1010,"name":"projectname2"}]";
json = json.substring(0, json.length() - 2); //strip the }]
String newJSON = json.concat(","age": 21}]"); // add in the new key and end
System.out.println(newJSON); // [{"id":1010,"name":"projectname2","age": 21}]

Answer 2

另一种方法来实现这一点(你说你正在使用JSONObject的现在，JSONObjects不是默认库)

取字符串

[{"id":1010,"name":"projectname2"}]

并将其转换为JSONObject

转换后，您可以使用JSONObject.append()将值为"21"的键"age"追加到它。