迭代遍歷一個對象遞歸查找孩子，沿途拾取父母，祖父母等

Question

我有一個包含菜單的對象。

我想輸入一個category ID並得到category name，然後向後移動以找到它的parents。在一個對象中這並不容易，所以我想要沿着這個方向趕上parents。

我遇到的問題是如何在resetparents未找到結尾的孩子，並且無處可去。

這就是我想：

var data = [ 
 
\t { 
 
     "tree_id": "10", 
 
     "name": "babies & children", 
 
     "parent": null, 
 
     "position": "1" 
 
    }, { 
 
     "tree_id": "2", 
 
     "name": "clothing", 
 
     "parent": null, 
 
     "position": "1", 
 
     "children": [{ 
 
      "tree_id": "15", 
 
      "name": "kids", 
 
      "parent": "2", 
 
      "position": "3", 
 
      "children": [{ 
 
       "tree_id": "78", 
 
       "name": "fourToTen", 
 
       "parent": "15", 
 
       "position": "3", 
 
       "children": [{ 
 
        "tree_id": "102", 
 
        "name": "fourToSix", 
 
        "parent": "78", 
 
        "position": "3" 
 
       }] 
 
      }] 
 
     }] 
 
    }, { 
 
     "tree_id": "55", 
 
     "name": "toys", 
 
     "parent": null, 
 
     "position": "1", 
 
     "children": [{ 
 
      "tree_id": "35", 
 
      "name": "lego", 
 
      "parent": "55", 
 
      "position": "3" 
 
     }] 
 
    } 
 
]; 
 

 
var crumbs = []; 
 
function getParts(data, elem) { 
 
    for(var i = 0; i < data.length; i++) { 
 
     var obj = data[i]; 
 
     if(obj.children !== undefined){ 
 
      /* push parent into crumbs */ 
 
      crumbs.push(obj.name); 
 
     \t if(obj.children[0].tree_id === elem){ 
 
      \t /* if we've found what we're looking, we're done */ 
 
       crumbs.push(obj.children[0].name); 
 
       console.log(crumbs); 
 
      } else { 
 
      \t /* reset parents */ 
 
       crumbs = []; /* <-- this is wrong here */ 
 
       /* not found, keep recursing */ 
 
      \t getParts(obj.children, elem); 
 
      } 
 
     } 
 
    } 
 
} 
 
/* I want this to return 
 
[ 
 
    "clothing", 
 
    "kids", 
 
    "fourToTen", 
 
    "fourToSix" 
 
] 
 

 
but it returns 
 
[ 
 
    "fourToTen", 
 
    "fourToSix" 
 
] 
 
*/ 
 
getParts(data, '102');

的問題是，我怎麼能保存parents陣列直到我在該行的結束和孩子找不到，然後重置它呢？

Here's a fiddle if that's your preferred playround

Answer 1

假設category id = tree_id和category_name = name

你需要沿途對待你data對象就像一棵樹，然後橫向它和跟蹤的父母。如果發現有問題，請轉儲您需要的信息。

所以data基本上是你將要橫穿的對象數組。

例子：

"use strict"; 
var data = [ 
    { 
     "tree_id": "10", 
     "name": "babies & children", 
     "parent": null, 
     "position": "1" 
    }, 
    { 
     "tree_id": "2", 
     "name": "clothing", 
     "parent": null, 
     "position": "1", 
     "children": [{ 
      "tree_id": "15", 
      "name": "kids", 
      "parent": "2", 
      "position": "3", 
      "children": [{ 
       "tree_id": "78", 
       "name": "fourToTen", 
       "parent": "15", 
       "position": "3", 
       "children": [{ 
        "tree_id": "102", 
        "name": "fourToSix", 
        "parent": "78", 
        "position": "3" 
       }] 
      }] 
     }] 
    }, 
    { 
     "tree_id": "55", 
     "name": "toys", 
     "parent": null, 
     "position": "1", 
     "children": [{ 
      "tree_id": "35", 
      "name": "lego", 
      "parent": "55", 
      "position": "3" 
     }] 
    } 
]; 

// Solution 
function transverse(root, tree, targetId) { 
    tree.push({ 
     catId : root.tree_id, 
     catName : root.name 
    }); 

    /* this if() must come first otherwise fails if you want to stop before end */ 
    if (root.tree_id === targetId) { 
     console.log("Found id:" + targetId+ ", name=" + root.name); 
     console.log("Dumping parent info => " + JSON.stringify(tree)); 
     return tree; 
    } 

    if (root.hasOwnProperty("children") && root.children instanceof Array) 
     root.children.forEach(child => { 
      transverse(child, tree, targetId); 
     }); 

} 

data.forEach(item => { 
    transverse(item, [], /*Looking for Id=*/"102"); 
}); 

console.log("done"); 

輸出：

Found id:102, name=fourToSix 
Dumping parent info => 
[ 
    {"catId":"2","catName":"clothing"}, 
    {"catId":"15","catName":"kids"}, 
    {"catId":"78","catName":"fourToTen"}, 
    {"catId":"102","catName":"fourToSix"}] 
]