Question

我正在使用骆驼弹簧，我将Json数组作为来自api的响应，如下所示：

[{"userName":"name","email":"email"}]

我的对象是这样的：

public class Response{

private String userName;
private String email;

setters & getters
}

我的路线创建者：

from("seda:rout")
.routeId("distinaation")
          .unmarshal()
          .json(JsonLibrary.Jackson, Request.class)
          .bean(Bean.class, "processRequest")
          .to(destination)
          .unmarshal()
          .json(JsonLibrary.Jackson, Response.class)
          .bean(Bean.class, "processResponse")

错误：

Can not desserialize instance of com.liena.Response out of START_ARRAY token

有什么方法可以直接将json数组解组到我的obj吗？

Answer 1

我通过在处理器中使用对象映射器解决了这个问题。

我的路线创建者：

from("seda:rout")
.routeId("distinaation")
          .unmarshal()
          .json(JsonLibrary.Jackson, Request.class)
          .bean(Bean.class, "processRequest")
          .to(destination)
          .bean(Bean.class, "processResponse")

处理器：

public Response processResponse(Exchange exchange) throws IOException {
        String responseStr = exchange.getIn().getBody().toString();
        ObjectMapper mapper = new ObjectMapper();
        List<Response> list = mapper.readValue(responseStr, new TypeReference<List<Response>>({});
        Response response = list.get(0);
        ...}

执行此操作的另一个简短方法是：在路径生成器中：

from("seda:rout")
.routeId("distinaation")
          .unmarshal()
          .json(JsonLibrary.Jackson, Request.class)
          .bean(Bean.class, "processRequest")
          .to(destination)
          .unmarshal(new ListJacksonDataFormat(Response.class))
          .bean(Bean.class, "processResponse")

处理器：

public Response processResponse(List<Response> list {
       Response response = list.get(0);
       ...
}

骆驼：解组json数组到pojo对象

1 个答案: