遞回回圈構建物件

我無法弄清楚如何正確解釋這一點，但我想制作一個采用此陣列的函式：

const tasks = [
    {id: 1, goal: 'Clean apartment', parent_id: null},
    {id: 2, goal: 'Clean bathroom', parent_id: 1},
    {id: 3, goal: 'Clean kitchen', parent_id: 1},
    {id: 4, goal: 'Clean sink', parent_id: 2},
    {id: 5, goal: 'Clean shower', parent_id: 2},
    {id: 6, goal: 'Make app', parent_id: null}

]

并生成此物件：

{id: 1, goal: 'Clean apartment', parent_id: null, children: [
    {id: 2, goal: 'Clean bathroom', parent_id: 1, children: [
        {id: 4, goal: 'Clean sink', parent_id: 2, children: []},
        {id: 5, goal: 'Clean shower', parent_id: 2, children: []},
    ]},
    {id: 3, goal: 'Clean kitchen', parent_id: 1, children: []},
]},
{id: 6, goal: 'Make app', parent_id: null, children: []}

編輯：到目前為止我做了這個，但它只回傳孩子的第一層：

function addChildren(tasks, id) {
    var task = tasks.find(task => task.id === id)
    var children = tasks.filter(task => task.parent_id === id)
    
    task.children = children
    return task
}

var newTask = addChildren(tasks, 1)
console.log(newTask)

編輯 2：我嘗試使函式遞回，但我收到一條錯誤訊息，說“tasks.find 不是函式”。

function addChildren(tasks, id) {
    var task = tasks.find(task => task.id === id)
    var children = tasks.filter(task => task.parent_id === id)
    
    task.children = children
    
    task.children.forEach(child => {
        addChildren(child, child.id)
    })
    return task
}

var newTask = addChildren(tasks, 1)
console.log(newTask)

uj5u.com熱心網友回復：

使函式遞回并為每個孩子呼叫它。

const tasks = [
    {id: 1, goal: 'Clean apartment', parent_id: null},
    {id: 2, goal: 'Clean bathroom', parent_id: 1},
    {id: 3, goal: 'Clean kitchen', parent_id: 1},
    {id: 4, goal: 'Clean sink', parent_id: 2},
    {id: 5, goal: 'Clean shower', parent_id: 2},
    {id: 6, goal: 'Make app', parent_id: null}

]

function addChildren(tasks, idOrTask) {
    var task = (typeof idOrTask === 'number') ? tasks.find(task => task.id === idOrTask) : idOrTask;
    var children = tasks.filter(t => t.parent_id === task.id);
    
    task.children = children.map(child => addChildren(tasks, child));
    return task
}

var newTask = addChildren(tasks, 1)
console.log(newTask)

我已經稍微優化了函式以避免不必要的呼叫find。

uj5u.com熱心網友回復：

您可以過濾與給定 id 匹配的那些，并使用遞回呼叫計算它們的子級：

const nest = (xs, parent = null) =>
  xs .filter (({parent_id}) => parent_id == parent)
     .map (({id, ...rest}) => ({id, ...rest, children: nest (xs, id)}))
 

const tasks = [{id: 1, goal: 'Clean apartment', parent_id: null}, {id: 2, goal: 'Clean bathroom', parent_id: 1}, {id: 3, goal: 'Clean kitchen', parent_id: 1}, {id: 4, goal: 'Clean sink', parent_id: 2}, {id: 5, goal: 'Clean shower', parent_id: 2}, {id: 6, goal: 'Make app', parent_id: null}]

console .log (nest (tasks))

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

如果您不再需要parent_id輸出中的 now-redundant，只需將map上面的行替換為：

    .map (({id, parent_id, ...rest}) => ({id, ...rest, children: nest (xs, id)}))

uj5u.com熱心網友回復：

const tasks = [
    {id: 1, goal: 'Clean apartment', parent_id: null},
    {id: 2, goal: 'Clean bathroom', parent_id: 1},
    {id: 3, goal: 'Clean kitchen', parent_id: 1},
    {id: 4, goal: 'Clean sink', parent_id: 2},
    {id: 5, goal: 'Clean shower', parent_id: 2},
    {id: 6, goal: 'Make app', parent_id: null}

]

function buildTree(root) {
    return {
      goal: root.goal, // Store the goal
      children: tasks
        .filter(task => task.parent_id === root.id) // All tasks with parent_id  = root.id
        .map(child => buildTree(child)) // Recursively collect their children
    }
}

const result = buildTree(tasks[0]) // Build the tree starting with 'Clean apartment'
console.log(result);

您可以通過不傳遞 ID 來稍微簡化邏輯

uj5u.com熱心網友回復：

為什么需要遞回？您可以使用這種迭代方式

function buildTree(tasks) {
  const tasksById = tasks.reduce((acc, task) => ({
    ...acc,
    [task.id]: { ...task }
  }), {})

  return Object.values(tasksById).reduce((tree, task) => {
    const parent = tasksById[task.parent_id]
    if (!parent) {
      return tree.concat(task)
    }

    parent.children = parent.children || []
    parent.children.push(task)
    return tree
  }, [])
}

標籤：javascript 递归

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