从json中递归查找依赖项

Question

data = [
    {
        "StoryID":"B-1",
        "Dependencies":[{"StoryID":"B-2",},{"StoryID":"B-3",}]
    },
    {
        "StoryID":"B-2",
        "Dependencies":[]
    },
    {
        "StoryID":"B-3",
        "Dependencies":[{"StoryID":"B-4",},{"StoryID":"B-5",}]
    },
    {
        "StoryID":"B-4",
        "Dependencies":[]
    },
    {
        "StoryID":"B-5",
        "Dependencies":[{"StoryID":"B-6"}]
    },
    {
        "StoryID":"B-6",
        "Dependencies":[]
    }
]

var storyID = "B-3";
get_d(data,storyID ,Dependencies = []);

function get_d(data,storyID ,Dependencies = []){
    temp = []
    for(i in data){
        if(storyID == data[i]['StoryID']){
            depend = data[i]['Dependencies'];
            for(j in depend){
                Dependencies.push(depend[j]["StoryID"]);
            }
        }
    }
}

我试图从上面的数据中找到特定故事的相关故事，我使用上面的函数只获得一个级别的数据，但我希望像这样递归输出，

{
  "B-3": [
    {
      "B-4": [

      ],
      "B-5": [
        {
          "Number": "B-6",
        },
        {
          "Number": "B-7",
        }
      ]
    }
  ]
}

Answer 1

您可以将数据转换为哈希表，并通过使用递归获得依赖项。

它仍然是相当不清楚，你想用树叶做什么。

​

function getDependencies(data, StoryID) {
    function getProperty(k) {
        return function (o) { return o[k]; };
    }

    function getD(r, StoryID) {
        r[StoryID] = object[StoryID].reduce(getD, {});
        return r;
    }

    var object = data.reduce(function (r, o) {
            r[o.StoryID] = (o.Dependencies || []).map(getProperty('StoryID'));
            return r;
        }, {});

    return [StoryID].reduce(getD, {});
}

var data = [{ StoryID: "B-1", Dependencies: [{ StoryID: "B-2" }, { StoryID: "B-3" }] }, { StoryID: "B-2", Dependencies: [] }, { StoryID: "B-3", Dependencies: [{ StoryID: "B-4" }, { StoryID: "B-5" }] }, { StoryID: "B-4", Dependencies: [] }, { StoryID: "B-5", Dependencies: [{ StoryID: "B-6" }] }, { StoryID: "B-6", Dependencies: [] }],
    result =  getDependencies(data, 'B-3')

console.log(result);

.as-console-wrapper { max-height: 100% !important; top: 0; }

​