當陣列中的新物件被更新時，函式不運行-防止物件陣列中的重復。

試圖改進這段代碼，防止推送()時物件陣列中的重復。

const mainArray = [{Id:1, Name: "test",SubDirectoryCount:10,Files:[]}】。]
const directories = [{Id:56, Name: "Client 2",SubDirectoryCount: 1,ParentId:1, Files: []},{Id:53, Name: "Client 3",SubDirectoryCount:1, ParentId: 1,Files:[]}, {Id: 50,Name:"Client 4", SubDirectoryCount: 1,ParentId:1, Files: []},{Id:48, Name: "Client 5",SubDirectoryCount:1, ParentId: 53,Files:[]}, {Id: 47,Name:"Client 5", SubDirectoryCount。 1,ParentId:53, Files: []},{Id:49, Name: "Client 5",SubDirectoryCount: 1,ParentId:53,Files:[]}】。]

//創建平面串列。
const tmp = mainArray.concat（directories）。
tmp.forEach(directory => {
  const parent = tmp.find(item =>/span> item. Id === directory.ParentId）。)

  if(parent) {
      parent.SubDirectories = parent.SubDirectories || [] 。
      parent.SubDirectories.push（目錄）。
  }
});

console.log(mainArray)

我的嘗試，但是，當目錄被更新為新物件時，它將不再構建。基本上，我只是在檢查它是否已經在陣列中，而不是推送。

const mainArray = [{Id:1, Name: "test",SubDirectoryCount:10,Files:[]}】。]
const directories = [{Id:56, Name: "Client 2",SubDirectoryCount: 1,ParentId:1, Files: []},{Id:53, Name: "Client 3",SubDirectoryCount:1, ParentId: 1,Files:[]}, {Id: 50,Name:"Client 4", SubDirectoryCount: 1,ParentId:1, Files: []},{Id:48, Name: "Client 5",SubDirectoryCount:1, ParentId: 53,Files:[]}, {Id: 47,Name:"Client 5", SubDirectoryCount。 1,ParentId:53, Files: []},{Id:49, Name: "Client 5",SubDirectoryCount: 1,ParentId:53,Files:[]}】。]

const tmp = mainArray.concat（directories）。
tmp.forEach(directory => {
  const parent = tmp.find(item =>/span> item. Id === directory.ParentId）。)

  if(parent) {
      parent.SubDirectories = parent.SubDirectories || [] 。
      const found = parent.SubDirectories. some(el => el.Id == directory.Id)
      if (!found) parent.SubDirectories.push（目錄）。
  }
});

console.log(mainArray)
    

在這行代碼中，是否有辦法使用include()并檢查是否存在而不添加到父級？

const parent = tmp. find(item => item.Id == directory.ParentId) 。

原來的問題

而且當你點擊另一個切換時，它會重復所有的東西

uj5u.com熱心網友回復：

簡單地說，就是

const directories = 
  [{ Id: 56, Name: 'Client 2', SubDirectoryCount: 1, ParentId: 1, Files: [] } 
  , { Id: 53, Name: 'Client 3', SubDirectoryCount: 1, ParentId: 1, Files: [] } 
  , { Id: 50, Name: 'Client 4', SubDirectoryCount: 1, ParentId: 1, Files: [] } 
  , { Id: 48, Name: 'Client 5', SubDirectoryCount: 1, ParentId: 53, Files: [] } 
  , { Id: 47, Name: 'Client 5', SubDirectoryCount: 1, ParentId: 53, Files: [] } 
  , { Id: 47, Name: 'Client 5', SubDirectoryCount: 1, ParentId: 53, Files: [] } 
  ] 

for (let index= directories.length; --index;)
  {
  if (index > directories.findIndex(x=>/span>x. Id====directories[index].Id)
    directories.splice(index,1)
  }
console.log( directories ) 

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

uj5u.com熱心網友回復：

你可以做到：

const directories = [{Id: 56, Name: 'Client 2', SubDirectoryCount: 1, ParentId: 1, Files: []}, {Id: 53, Name: 'Client 3', SubDirectoryCount: 1, ParentId: 1, Files: []}, {Id: 50, Name: 'Client 4', SubDirectoryCount: 1, ParentId: 1, Files: []}, {Id: 48, Name: 'Client 5', SubDirectoryCount: 1, ParentId: 53, Files: []}, {Id: 47, Name: 'Client 5', SubDirectoryCount: 1, ParentId: 53, Files: []}, {Id: 47, Name: 'Client 5', SubDirectoryCount: 1, ParentId: 53, Files: []}]

const result = Object.values（directories。 reduce((a, c) =>（a[c. Id] = a[c.Id] || c, a), {})

console.log（result）

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