基于给定搜索项的未知深度嵌套对象数组的过滤

Question

这里有类似的答案，但我看到和测试过的所有答案都没有达到两个层次，所以我不认为这是一个重复的问题.我正在尝试过滤一个对象数组。每个对象都可以有其他深度未知的嵌套对象。对于我的数据，treeView是treeData的集合，treeNode(子)将具有相同的接口。

export interface TreeView = TreeData[]

export interface TreeData {
    label: string;
    children?: TreeData[]
}

const treeData = 
[
  {label: 'obj-1', children: [ {label: 'bananas'},{label: 'apples', children: [
       {label: 'bananas',},{label: 'apples'}, {label: 'coconut',  children: [
           {label: 'bananas'},{label: 'oranges',  },
]},
]}]},
  
{label: 'obj-2', children: [ {label: 'apples'},{label: 'apples', children: [
       {label: 'oranges',},{label: 'apples'}, {label: 'coconut',  children: [
           {label: 'bananas'},{label: 'oranges', children: [
             {label: 'bananas'},{label: 'oranges',  },
] },
]},
]}]},

  {label: 'obj-3', children: [ {label: 'apples'},{label: 'mango'},{label: 'apples', children: [
       {label: 'oranges',},{label: 'apples'}, {label: 'coconut',  children: [
           {label: 'bananas'},{label: 'oranges', children: [
             {label: 'bananas'},{label: 'mango',  },
] },
]},
]}]},
]

应该删除带有与searchTerm不匹配的标签的对象，如果对象具有子属性，则重复该过程。我一直试图用递归来解决这个问题。这是我的密码。

const recurse = (array, searchTerm) => {
  return array.map((element) => {
  return {...element, subElements: element.subElements.filter((subElement) => subElement.label === searchTerm)}
})
}

const filterArray = (myArray,value) => {
  for(let i = 0; i < myArray.length;i++){
    if(myArray.children){
      if(myArray[i].children){
          filterArray(myArray[i].children,value) 
        }
    }
    } 
     recurse(myArray,value)
  }


console.log(filterArray(treeData,'bananas'))

在上面的示例中，搜索‘香蕉’将尽可能深地过滤整个数组，删除标签支柱不等于香蕉的任何对象；如果一个对象没有子对象，它的标签支柱是否等于搜索项，如果不是从其父数组中删除它。提前谢谢你！

我想返回原始array.Only对象的过滤版本，其标签将显示为“香蕉”

​

Answer 1

我想你会想要这样的东西：

const filterByLabel = (array, searchTerm) => {
    return array.reduce((prev, curr) => {
        const children = curr.children ? filterByLabel(curr.children, searchTerm) : undefined;
        
        return curr.label === searchTerm || children?.length > 0 ? [...prev, { ...curr, children }] : prev;
    }, []);
}

filterByLabel(treeData, 'bananas');