Question

我需要从JSON构建一个树。对于每个restProjectLevel，JSON结构都会读取：restProjectLevels我有一个id = level。 sequenceNr = root或child。对于每个restProject：restProjects我有一个id = projectId和一个projectIdHook =附加的projectId。

我认为JSON数据最能解释它。

{
    "id": 3,
    "description": "New Project Plan",
    "restProjectLevels": [
        {
            "id": 19,
            "sequenceNr": 0,
            "restProjects": [
                {
                    "id": 28,
                    "projectName": "Project A",
                    "description": "",
                    "projectLevelId": 19,
                    "projectIdHook": 0,
                    "restProjectProcesses": []
                },
                {
                    "id": 29,
                    "projectName": "Project B",
                    "description": "",
                    "projectLevelId": 19,
                    "projectIdHook": 0,
                    "restProjectProcesses": []
                },
                {
                    "id": 30,
                    "projectName": "Project C",
                    "description": "",
                    "projectLevelId": 19,
                    "projectIdHook": 0,
                    "restProjectProcesses": []
                },
                {
                    "id": 41,
                    "projectName": "New",
                    "description": "",
                    "projectLevelId": 19,
                    "projectIdHook": 0,
                    "restProjectProcesses": []
                }
            ]
        },
        {
            "id": 20,
            "sequenceNr": 1,
            "restProjects": [
                {
                    "id": 31,
                    "projectName": "Project A.1",
                    "description": "",
                    "projectLevelId": 20,
                    "projectIdHook": 28,
                    "restProjectProcesses": []
                },
                {
                    "id": 33,
                    "projectName": "Project B.1",
                    "description": "",
                    "projectLevelId": 20,
                    "projectIdHook": 29,
                    "restProjectProcesses": []
                },
                {
                    "id": 35,
                    "projectName": "Project C.1",
                    "description": "",
                    "projectLevelId": 20,
                    "projectIdHook": 30,
                    "restProjectProcesses": []
                }
            ]
        },
        {
            "id": 21,
            "sequenceNr": 2,
            "restProjects": [
                {
                    "id": 32,
                    "projectName": "Project A.2",
                    "description": "",
                    "projectLevelId": 21,
                    "projectIdHook": 31,
                    "restProjectProcesses": []
                },
                {
                    "id": 36,
                    "projectName": "Project C.2",
                    "description": "",
                    "projectLevelId": 21,
                    "projectIdHook": 35,
                    "restProjectProcesses": []
                }
            ]
        },
        {
            "id": 22,
            "sequenceNr": 3,
            "restProjects": [
                {
                    "id": 34,
                    "projectName": "Projet B.4",
                    "description": "",
                    "projectLevelId": 22,
                    "projectIdHook": 33,
                    "restProjectProcesses": []
                },
                {
                    "id": 37,
                    "projectName": "Project C.3",
                    "description": "",
                    "projectLevelId": 22,
                    "projectIdHook": 36,
                    "restProjectProcesses": []
                }
            ]
        },
        {
            "id": 23,
            "sequenceNr": 4,
            "restProjects": []
        }
    ]
}

Answer 1

确定解决了这个问题！不是最优雅的解决方案，但它有效...

    for(RestProjectLevel projectLevel: projectLevelList) {
        if(projectLevel.getSequenceNr() == 0) {
            for(RestProject project : projectLevel.getRestProjects()) {
                restParentNode = new RestParentNode();
                traverse(sortedRestProjects, project, restParentNode);
                restTree.nodes.add(restParentNode);
            }
        }
    }

private void traverse(List<RestProject> restProjects, RestProject project, RestParentNode restParentNode) {
    for (RestProject restProject : restProjects) {
        if(project.getId() == restProject.getProjectIdHook() || project.getId() == restProject.getId()) {

            String projectName = restProject.getProjectName();

            if(hasNode(projectName)) {
                continue;
            } else {
                RestNode restNode = new RestNode();
                restNode.nodes.add(restProject);
                restParentNode.nodes.add(restNode);
                traverse(restProjects, restProject, restParentNode);
            }
        }
    }
}

谢谢！

在Java中将JSON数据结构遍历到树中

1 个答案: