[英]Filter N level nested array of objects based upon multiple properties
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'}
]
}
]
}
];
Problem:问题:
I need to perform a keyword search on the question, and if a question is found at a node - 3 let's say then we need to return that node and all the parent nodes of that object.我需要对问题执行关键字搜索,如果在节点处找到问题 - 比方说 3,那么我们需要返回该节点和该 object 的所有父节点。
For example, if I search for 'hello there', the final tree should be: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'}
]
}
]
}
];
We can have children or questions [] at any node.我们可以在任何节点有孩子或问题 []。
I am able to find the questions matching the search string, but I am not able to remove the unwanted nodes from the tree.我能够找到与搜索字符串匹配的问题,但无法从树中删除不需要的节点。 Here is the code for that: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)
This should work for you.这应该适合你。 The demo code is in stackblitz.演示代码在 stackblitz 中。 Check the result in the console.在控制台中查看结果。
searchString = 'hello';
filteredData = [];
ngOnInit(): void {
this.filteredData = [];
this.data.forEach(node => {
if (this.checkQtn(node)) {
this.filteredData.push(node);
}
});
console.log(this.filteredData);
}
checkQtn(node): Boolean {
let response: Boolean = false;
if (node.questions) {
let qtns = [];
qtns = node.questions;
qtns.forEach(el => {
const eachQtn: string = el.name;
if (eachQtn.includes(this.searchString)) {
response = true;
}
});
}
if (!response && node.children) {
for (let i = 0; i < node.children.length && !response; i++) {
response = this.checkQtn(node.children[i]);
}
}
return response;
}
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