我真的不知道这个。我正在尝试展平属于特定节点的子级category_id
。
var categories = [{
"category_id": "66",
"parent_id": "59"
}, {
"category_id": "68",
"parent_id": "67",
}, {
"category_id": "69",
"parent_id": "59"
}, {
"category_id": "59",
"parent_id": "0",
}, {
"category_id": "67",
"parent_id": "66"
}, {
"category_id": "69",
"parent_id": "59"
}];
或视觉上:
我最接近的是递归遍历找到的第一项:
function children(category) {
var children = [];
var getChild = function(curr_id) {
// how can I handle all of the cats, and not only the first one?
return _.first(_.filter(categories, {
'parent_id': String(curr_id)
}));
};
var curr = category.category_id;
while (getChild(curr)) {
var child = getChild(curr).category_id;
children.push(child);
curr = child;
}
return children;
}
children(59)
的当前输出为['66', '67', '68']
。预期输出为
['66', '67', '68', '69']
最佳答案
我没有测试,但应该可以:
function getChildren(id, categories) {
var children = [];
_.filter(categories, function(c) {
return c["parent_id"] === id;
}).forEach(function(c) {
children.push(c);
children = children.concat(getChildren(c.category_id, categories));
})
return children;
}
我正在使用lodash。
编辑:我测试了它,现在应该可以了。参见缩略语:https://plnkr.co/edit/pmENXRl0yoNnTczfbEnT?p=preview
您可以通过丢弃过滤后的类别来进行一个小的优化。
function getChildren(id, categories) {
var children = [];
var notMatching = [];
_.filter(categories, function(c) {
if(c["parent_id"] === id)
return true;
else
notMatching.push(c);
}).forEach(function(c) {
children.push(c);
children = children.concat(getChildren(c.category_id, notMatching));
})
return children;
}
关于javascript - Javascript扁平化深层嵌套的子级,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/36702379/