Question

我有一些代码可以将restful响应转换为递归结构。但是，我想要解析为树的内容被包装到“treeprop”属性中。是否有更方便的方法来解析真实内容？

ObjectMapper mapper = new ObjectMapper();
JsonFactory jfac = new JsonFactory();
JsonParser jp = jfac.createParser(inputStream);
JsonNode rootNode = mapper.readTree(jp);

JsonNode path = rootNode.path("treeprop");
String realContent = mapper.writeValueAsString(path);

MyTree mt = mapper.readValue(realContent, MyTree.class);
inputStream.close();

请注意，解析本身不是问题所在。上面的代码确实将inputStream正确转换为树。但是，json经常大于1MB，所以临时将它存储在String中是不能运行时效率的。

{
  "treeprop": {
    "id": 0,
    "children": [
      {
        "id": 2,
        "children": [
          {
            "id": 3,
            "children": []
          },
          {
            "id": 4,
            "children": []
          }
        ]
      },
      {
        "id": 1,
        "children": []
      }
    ]
  }
}

课程看起来大致像这样

class MyTree {
   public Integer id;
   public List<MyTree> children;
}

所以真正的问题是：是否有更有效的替代方案可以实现相同的目标：

JsonNode path = rootNode.path("treeprop");
String realContent = mapper.writeValueAsString(path);
MyTree mt = mapper.readValue(realContent, MyTree.class);

Answer 1

public class MyTreeWrapper {
    private MyTree treeprop;

    // getter, setter
}

...

ObjectMapper mapper = new ObjectMapper();
MyTreeWrapper wrapper = mapper.readValue(inputStream, MyTreeWrapper.class);
MyTree tree = wrapper.getTreeprop();

替代：

MyTree tree = mapper.reader(MyTree.class).readValue(rootNode.path("treeprop"));

如何在使用jackson解析时防止冗余字符串转换步骤？

1 个答案: