深度嵌套的对象数组，查找并删除具有特定 ID 的 Object

Question

我正在尝试通过 ID 查找 Object 并将其删除。 （当然，如果这个 object 有子任务，那么挂在这棵树上的所有对象也会被删除）。 我有一个这样的结构（例如），它可以变得非常大和深：

[
  {
    "id": 0,
    "lane": 0,
    "name": "Task A",
    "start": 0,
    "end": 10000,
    "subTasks": [
      {
        "id": "0.1",
        "lane": 0,
        "name": "Subtask",
        "start": 0,
        "end": 10000,
        "class": "danger",
        "sublane": 0,
        "subTasks": [
          {
            "id": "0.1.1",
            "name": "Subtask",
            "start": 0,
            "end": 10000,
            "subTasks": [
              {
                "id": "0.1.1.1",
                "name": "Subtask",
                "start": 0,
                "end": 10000,
                "subTasks": [
                  {
                    "id": "0.1.1.1.1",
                    "name": "Subtask",
                    "start": 0,
                    "end": 10000
                  },
                  {
                    "id": "0.1.1.1.2",
                    "name": "Subtask",
                    "start": 0,
                    "end": 10000
                  }
                ]
              },
              {
                "id": "0.1.1.2",
                "name": "Subtask",
                "start": 0,
                "end": 10000
              }
            ]
          },
          {
            "id": "0.1.2",
            "name": "Subtask",
            "start": 0,
            "end": 10000
          },
          {
            "id": "0.1.3",
            "name": "Subtask",
            "start": 0,
            "end": 10000
          }
        ]
      },
      {
        "id": "0.2",
        "name": "Subtask",
        "start": 0,
        "end": 10000
      }
    ],
    "class": "danger",
    "sublane": 0
  },
  {
    "id": 1,
    "lane": 2,
    "name": "Task B",
    "start": 15000,
    "end": 25000,
    ],
    "class": "success",
    "sublane": 0
  }
]

现在我想删除 ID = 0.1.1.1.1 例如，但它应该以同样的方式与其他所有 Object 一起工作，无论它嵌套多深。

使用此 dfs 算法查找和编辑 Im：

   edit: function (name, start, end) {
        for (let obj of gantt.items()) {
            result = dfs(obj, id);
            if (result) {
                result.name = name;
                result.start = start;
                result.end = end;
            }
        }
    
      dfs: function (obj, targetId) {
            if (obj.id === targetId) {
                return obj;
            }
            if (obj.subTasks) {
                for (let item of obj.subTasks) {
                    let check = dfs(item, targetId);
                    if (check) {
                        return check;
                    }
                }
            }
            return null;
        },

但是如何删除/删除特定的 Object？

Answer 1

递归过滤器的种类：

 const data = [{"id":"0","lane":0,"name":"Task A","start":0,"end":10000,"Subtask":[{"id":"0.1","lane":0,"name":"Subtask","start":0,"end":10000,"class":"danger","sublane":0,"Subtask":[{"id":"0.1.1","name":"Subtask","start":0,"end":10000,"Subtask":[{"id":"0.1.1.1","name":"Subtask","start":0,"end":10000,"Subtask":[{"id":"0.1.1.1.1","name":"Subtask","start":0,"end":10000},{"id":"0.1.1.1.2","name":"Subtask","start":0,"end":10000}]},{"id":"0.1.1.2","name":"Subtask","start":0,"end":10000}]},{"id":"0.1.2","name":"Subtask","start":0,"end":10000},{"id":"0.1.3","name":"Subtask","start":0,"end":10000}]},{"id":"0.2","name":"Subtask","start":0,"end":10000}],"class":"danger","sublane":0},{"id":"1","lane":2,"name":"Task B","start":15000,"end":25000,"class":"success","sublane":0}]; const removeById = (arr, targetId) => arr.reduce((acc, obj) => (obj.id === targetId)? acc: [...acc, {...obj, ...(obj.Subtask && { Subtask: removeById(obj.Subtask, targetId) }) } ], []); console.log(removeById(data, "0.1"));

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