如题，有时候需要对树形数组深层去找符合字段的那一串json，苦于循环找太费劲，索引选择扁平化，找起来方便很多
let treeList = [{
id: '1',
name: '水果',
value: 3,
children: [{
id: '1-1',
name: '西瓜',
value: 2,
children: [{
id: '1-1-1',
name: '西瓜皮',
value: 1,
children: [],
},
{
id: '1-1-2',
name: '西瓜种子',
value: 1,
children: [],
}
]
}, {
id: '1-2',
name: '柚子',
value: 1,
children: [],
}, {
id: '1-3',
name: '橘子',
value: 1,
children: [],
}]
},
{
id: '2',
name: '肉',
value: 4,
children: [{
id: '2-1',
name: '牛肉',
value: 1,
children: [{
id: '2-1-2',
name: '和牛肉',
value: 1,
children: [],
},
{
id: '2-1-2',
name: '澳牛肉',
value: 1,
children: [],
}
],
}, {
id: '2-2',
name: '羊肉',
value: 1,
children: [],
}, {
id: '2-3',
name: '猪肉',
value: 1,
children: [],
},
{
id: '2-4',
name: '鱼肉',
value: 1,
children: [],
}
]
},
{
id: '3',
name: '蔬菜',
value: 2,
children: [{
id: '3-1',
name: '大白菜',
value: 1,
children: [],
}, {
id: '3-2',
name: '菠菜',
value: 1,
children: [],
}]
},
];
console.log(treeList);
封装函数如下：
//扁平化 /* * bossArr 树形数据 * children 树形数据子数据的属性名, 本项目为 children * */ function extractTree(bossArr, children) { //如果为空 返回空（防止 children 递归报错） if (!Array.isArray(bossArr) && !bossArr.length) return []; let list = []; const getObj = (arr) => { arr.forEach((row) => { let obj = {}; obj = JSON.parse(JSON.stringify(row)); list.push(obj); if (row[children]) { getObj(row[children]); } }); return list; }; return getObj(bossArr); } const newList = extractTree(treeList, 'children'); console.log(newList);