【发布时间】:2020-11-07 02:18:33
【问题描述】:
data = [
{
name: "Parent Level 1",
questions: [
{
name: "question 1"
}
],
children: [
{
name: "Child 1 - P1",
questions: [
{
name: "ability to code"
},
{
name: "ability to do something"
}
],
children: [
{
name: "Child -2 P1",
questions: [
{
name: "figure out"
}
]
}
]
}
]
},
{
name : 'Parent Level 2',
questions : [
{name : 'question 1 P-2'}
]
},
{
name : 'Parent Level 3',
children: [
{
name : 'Child Level -1 P-3',
children: [
{
name : 'Child Level 2- P-3',
questions : [
{
name : 'Question level 2
}
]
}
]
questions: [
{name : 'hello there'}
]
}
]
}
];
问题:
我需要对问题执行关键字搜索,如果在某个节点找到问题 - 假设是 3,那么我们需要返回该节点以及该对象的所有父节点。
例如,如果我搜索“你好”,最终的树应该是:
[
{
name : 'Parent Level 3',
children: [
{
name : 'Child Level -1 P-3',
children: [
{
name : 'Child Level 2- P-3',
questions : []
}
]
questions: [
{name : 'hello there'}
]
}
]
}
];
我们可以在任何节点有子节点或问题 []。
我能够找到与搜索字符串匹配的问题,但我无法从树中删除不需要的节点。这是代码:
searchNode (data) {
for (let d of data) {
this.search(d)
}
}
search(data) {
let search = 'ability'
if(!!data.questions && data.questions.length > 0) {
data.questions = data.questions.filter((question) => {
return question.name.includes(search)
})
}
if(data.children && data.children.length > 0) {
searchNode(data.children)
}
}
search(data)
【问题讨论】:
标签: javascript angular lodash