Question

我想在以下代码片段上用Jackson替换JSON.simple：

JSONObject request = new JSONObject();
request.put("String key", /String value/);
request.put("String key", /int value/);
...

看起来像这样：

ObjectMapper mapper = new ObjectMapper();
JsonNode request = mapper.createObjectNode();
((ObjectNode) request).put("String key", /String value/);
((ObjectNode) request).put("String key", /int value/);

我发现强制转换和多余的声明有点过于复杂和丑陋。我有做错什么建议吗？

（我想通过REST实体发送此JSON）

Answer 1

使用ObjectNode代替JsonNode。试试这个：

    ObjectNode request = mapper.createObjectNode();
    request.put("key", "val");
    System.out.println(request.toString());

Answer 2

此链接提供了很好的信息。我觉得您应该拥有对象，而不是手动构建json。

https://www.baeldung.com/jackson-object-mapper-tutorial

public class Car {

    private String color;
    private String type;

    // standard getters setters
}

然后

ObjectMapper objectMapper = new ObjectMapper();
Car car = new Car("yellow", "renault");
String output = objectMapper.writeValueAsString(car);

输出：

{"color":"yellow","type":"renault"}

Answer 3

您的代码应该是这样

            ObjectMapper mapper = new ObjectMapper();
            ObjectNode request = mapper.createObjectNode();
            request.put("String key", /String value/);
            request.put("String key", /int value/);

用Jackson替换JSON.simple

3 个答案: