【发布时间】：2022-01-23 05:04:18
【问题描述】：
我正在编写一个递归函数，它需要在具有任何深度级别的对象数组中运行。 （如果它找到一个数组，它会在完成对象属性后运行到这个数组中）
我们的想法是在网页中创建一个通用表格，该表格可以处理任何类型的对象结构并呈现尊重其层次结构的元素。
我可以更深入任何级别，但它永远不会完成循环：
let keys = [
'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'filhos', 'netos', 'bisnetos'
];
let tempArr = [];
let counter = 0;
function renderer(arr) {
for (let x = 0; x < arr.length; x++) {
const currItem = arr[x];
for (let y = 0; y < keys.length; y++) {
const inner = currItem[keys[y]]
if (inner instanceof Array) {
tempArr = inner;
}
if (inner && !(inner instanceof Array)) {
console.log(`renderizando ${counter} camada: `, inner);
}
if (y === keys.length - 1) {
if (tempArr.length > 0) {
const children = tempArr;
tempArr = [];
return renderer(children);
} else {
continue;
}
}
}
}
counter++;
console.log('counter: ', counter);
return counter;
}
const data = [{
a: '1st item',
b: '2nd item',
c: '3rd item',
d: '4th item',
filhos: [{
a: 'filho 1st item',
b: 'filho 2nd item',
c: 'filho 3rd item',
d: 'filho 4th item',
netos: [{
a: 'neto 1st item',
b: 'neto 2nd item',
c: 'neto 3rd item',
d: 'neto 4th item',
bisnetos: [{
a: 'bisneto 1st item',
b: 'bisneto 2nd item',
c: 'bisneto 3rd item',
d: 'bisneto 4th item',
f: 'bisneto 5th item',
g: 'bisneto 6th item',
h: 'bisneto last item'
}],
f: 'neto 5th item',
g: 'neto 6th item',
h: 'neto last item'
}],
f: 'filho 5th item',
g: 'filho 6th item',
h: 'filho last item'
}],
f: '5th item',
g: '6th item',
h: 'last item'
},
{
a: '1st item',
b: '2nd item',
c: '3rd item',
d: '4th item',
filhos: [{
a: 'filho 1st item',
b: 'filho 2nd item',
c: 'filho 3rd item',
d: 'filho 4th item',
netos: [{
a: 'neto 1st item',
b: 'neto 2nd item',
c: 'neto 3rd item',
d: 'neto 4th item',
bisnetos: [{
a: 'bisneto 1st item',
b: 'bisneto 2nd item',
c: 'bisneto 3rd item',
d: 'bisneto 4th item',
f: 'bisneto 5th item',
g: 'bisneto 6th item',
h: 'bisneto last item'
}],
f: 'neto 5th item',
g: 'neto 6th item',
h: 'neto last item'
}],
f: 'filho 5th item',
g: 'filho 6th item',
h: 'filho last item'
}],
f: '5th item',
g: '6th item',
h: 'last item'
},
{
a: '1st item',
b: '2nd item',
c: '3rd item',
d: '4th item',
filhos: [{
a: 'filho 1st item',
b: 'filho 2nd item',
c: 'filho 3rd item',
d: 'filho 4th item',
netos: [{
a: 'neto 1st item',
b: 'neto 2nd item',
c: 'neto 3rd item',
d: 'neto 4th item',
bisnetos: [{
a: 'bisneto 1st item',
b: 'bisneto 2nd item',
c: 'bisneto 3rd item',
d: 'bisneto 4th item',
f: 'bisneto 5th item',
g: 'bisneto 6th item',
h: 'bisneto last item'
}],
f: 'neto 5th item',
g: 'neto 6th item',
h: 'neto last item'
}],
f: 'filho 5th item',
g: 'filho 6th item',
h: 'filho last item'
}],
f: '5th item',
g: '6th item',
h: 'last item'
},
]
renderer(data);看到它在第一个列表中的第一次迭代之后结束，而没有遇到接下来的两个对象。
有什么见解吗？
谢谢。
【问题讨论】：
-
return退出整个函数，结束内循环和外循环。也许你只是想跳出内循环？ -
@Barmar 我删除了所有退货并且它有效。我认为递归函数中的 return 关键字需要返回到以前的调用。您要回复一个答案然后我可以标记为已接受吗？谢谢！
-
如果递归调用的返回值为本层的返回值，则只需要返回即可。
标签： javascript arrays recursion