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如何使用 vue/javascript 通过多个属性过滤深度嵌套的 json

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慕桂英4014372 2024-01-18 15:52:17

我有一些具有以下结构的 JSON：{ "root": { "Europe": { "children": [ { "name": "Germany" }, { "name": "England", "children": [ { "name": "London", "search_words": ["city", "capital"], "children": [ { "name": "Westminster", "search_words": ["borough"] } ] }, { "name": "Manchester", "search_words": ["city"] } ] }, { "name": "France", "children": [ { "name": "Paris", "search_words": ["city", "capital"] } ] } ] }, "North America": { "children": [ { "name": "Canada", "children": [ { "name": "Toronto" }, { "name": "Ottawa" } ] }, { "name": "United States" } ] } }}我想根据文本搜索过滤 JSON。我应该能够通过 thename和 any进行搜索search_words。最后一个问题是 JSON 可以是任意深度，因此它需要能够搜索所有级别。我还需要循环遍历并打印出 HTML 中的 JSON（使用 Vue），并根据搜索进行更新。目前还不清楚如何在不知道 JSON 深度的情况下执行此操作？任何帮助将不胜感激！

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3 回答

holdtom

TA贡献1805条经验获得超10个赞

在开始之前，我们必须首先解决输入数据的不规则形状 -

const data2 =

{ name:"root"

, children:

Array.from

( Object.entries(data.root)

, ([ country, _ ]) =>

Object.assign({ name:country }, _)

)

}

console.log(JSON.stringify(data2, null, 2))

现在我们看到的data2是一个统一的{ name, children: [ ... ]}形状——

{

"name": "root",

"children": [

{

"name": "Europe",

"children": [

{ "name": "Germany" },

{

"name": "England",

"children": [

{

"name": "London",

"search_words": [ "city", "capital" ],

"children": [

{

"name": "Westminster",

"search_words": [ "borough" ]

}

]

{

"name": "Manchester",

"search_words": [ "city" ]

}

]

{

"name": "France",

"children": [

{

"name": "Paris",

"search_words": [ "city", "capital" ]

}

]

}

]

{

"name": "North America",

"children": [

{

"name": "Canada",

"children": [

{ "name": "Toronto" },

{ "name": "Ottawa" }

]

{ "name": "United States" }

]

}

]

}

现在我们编写一个通用的深度优先遍历函数，dft-

function* dft (t, path = [])

{ for (const _ of t.children ?? [])

yield* dft(_, [...path, t.name ])

yield [path, t]

}

我们的dft函数为我们的输入树中的path每个元素提供了一个-et

["root","Europe"]

{"name":"Germany"}

["root","Europe","England","London"]

{name:"Westminster", search_words:["borough"]}

["root","Europe","England"]

{name:"London", search_words:["city","capital"], children:[...]}

["root","Europe","England"]

{name:"Manchester", search_words:["city"]}

["root","Europe"]

{name:"England", children:[...]}

["root","Europe","France"]

{name:"Paris", search_words:["city","capital"]}

["root","Europe"]

{name:"France", children:[...]}

["root"]

{name:"Europe", children:[...]}

["root","North America","Canada"]

{name:"Toronto"}

现在我们知道了每个节点的路径，我们可以创建一个index使用path和 anysearch_words链接回该节点的路径 -

const index = t =>

Array.from

( dft(t)

, ([path, e]) =>

[ [...path, e.name, ...e.search_words ?? [] ] // all words to link to e

, e // e

]

)

.reduce

( (m, [ words, e ]) =>

insertAll(m, words, e) // update the index using generic helper

, new Map

)

这取决于通用助手insertAll-

const insertAll = (m, keys, value) =>

keys.reduce

( (m, k) =>

m.set(k, [ ...m.get(k) ?? [], value ])

, m

)

完成后index，我们有一种方法可以为任何搜索词创建快速查找 -

const myIndex =

index(data2)

console.log(myIndex)

Map

{ "Europe" =>

[{"name":"Germany"},{"name":"Westminster",...},{"name":"London",...},{"name":"Manchester",...},{"name":"England"...},{"name":"Manchester",...}]},{"name":"Paris",...},{"name":"France"...},{"name":"Europe"...},{"name":"Manchester",...}]},{"name":"France"...}]}]

, "Germany" =>

[{"name":"Germany"}]

, "England" =>

[{"name":"Westminster",...},{"name":"London",...},{"name":"Manchester",...},{"name":"England"...},{"name":"Manchester",...}]}]

, "London" =>

[{"name":"Westminster",...},{"name":"London",...}]

, "Westminster" =>

[{"name":"Westminster",...}]

, "borough" =>

[{"name":"Westminster",...}]

, "city" =>

[{"name":"London",...},{"name":"Manchester",...},{"name":"Paris",...}]

, "capital" =>

[{"name":"London",...},{"name":"Paris",...}]

, "Manchester" =>

[{"name":"Manchester",...}]

, "France" =>

[{"name":"Paris",...},{"name":"France"...}]

, "Paris" =>

[{"name":"Paris",...}]

, "North America" =>

[{"name":"Toronto"},{"name":"Ottawa"},{"name":"Canada"...},{"name":"United States"},{"name":"North America"...},

{"name":"United States"}]}]

, "Canada" =>

[{"name":"Toronto"},{"name":"Ottawa"},{"name":"Canada"...}]

, "Toronto" =>

[{"name":"Toronto"}]

, "Ottawa" =>

[{"name":"Ottawa"}]

, "United States" =>

[{"name":"United States"}]

}

这应该突出显示数据中剩余的不一致之处。例如，您有一些嵌套在city、capital或下的节点borough。另外值得注意的是，我们可能应该使用s.toLowerCase()所有索引键，以便查找不区分大小写。这是留给读者的练习。

创建index很容易，您只需要做一次-

const myIndex =

index(data2)

您的索引可以根据需要重复用于任意多次查找 -

console.log(myIndex.get("Toronto") ?? [])

console.log(myIndex.get("France") ?? [])

console.log(myIndex.get("Paris") ?? [])

console.log(myIndex.get("Canada") ?? [])

console.log(myIndex.get("Zorp") ?? [])

[{"name":"Toronto"}]

[{"name":"Paris",...},{"name":"France"...}]

[{"name":"Paris",...}]

[{"name":"Toronto"},{"name":"Ottawa"},{"name":"Canada"...}]

[]

将结果插入 Vue 应用程序由您完成。

反对回复 2024-01-18

江户川乱折腾

TA贡献1851条经验获得超5个赞

不完全清楚您要从问题中寻找什么，但我猜您需要修改数据以确保在渲染时正确突出显示匹配的数据？

这是使用对象扫描查找匹配对象的解决方案

// const objectScan = require('object-scan');

const myData = { root: { Europe: { children: [{ name: 'Germany' }, { name: 'England', children: [{ name: 'London', search_words: ['city', 'capital'], children: [{ name: 'Westminster', search_words: ['borough'] }] }, { name: 'Manchester', search_words: ['city'] }] }, { name: 'France', children: [{ name: 'Paris', search_words: ['city', 'capital'] }] }] }, 'North America': { children: [{ name: 'Canada', children: [{ name: 'Toronto' }, { name: 'Ottawa' }] }, { name: 'United States' }] } } };

// eslint-disable-next-line camelcase

const mySearchFn = (term) => ({ name, search_words = [] }) => name === term || search_words.includes(term);

const search = (input, searchFn) => objectScan(['**[*]'], {

filterFn: ({ value, context }) => {

if (searchFn(value)) {

const { children, ...match } = value;

context.push(match);

}

})(input, []);

console.log(search(myData, mySearchFn('Toronto')));

// => [ { name: 'Toronto' } ]

console.log(search(myData, mySearchFn('borough')));

// => [ { name: 'Westminster', search_words: [ 'borough' ] } ]

console.log(search(myData, mySearchFn('capital')));

// => [ { name: 'Paris', search_words: [ 'city', 'capital' ] }, { name: 'London', search_words: [ 'city', 'capital' ] } ]

.as-console-wrapper {max-height: 100% !important; top: 0}

这就是您如何注入信息，然后渲染管道可以获取这些信息

// const objectScan = require('object-scan');

// eslint-disable-next-line camelcase

const mySearchFn = (term) => ({ name, search_words = [] }) => name === term || search_words.includes(term);

const search = (input, searchFn) => objectScan(['**[*]'], {

filterFn: ({ value }) => {

if (searchFn(value)) {

value.css = { highlight: true };

return true;

} else {

delete value.css;

return false;

}

rtn: 'count' // return number of matches

})(input);

console.log(search(myData, mySearchFn('Toronto')));

// => 1

console.log(myData);

// => { root: { Europe: { children: [ { name: 'Germany' }, { name: 'England', children: [ { name: 'London', search_words: [ 'city', 'capital' ], children: [ { name: 'Westminster', search_words: [ 'borough' ] } ] }, { name: 'Manchester', search_words: [ 'city' ] } ] }, { name: 'France', children: [ { name: 'Paris', search_words: [ 'city', 'capital' ] } ] } ] }, 'North America': { children: [ { name: 'Canada', children: [ { name: 'Toronto', css: { highlight: true } }, { name: 'Ottawa' } ] }, { name: 'United States' } ] } } }

console.log(search(myData, mySearchFn('borough')));

// => 1

console.log(myData);

// => { root: { Europe: { children: [ { name: 'Germany' }, { name: 'England', children: [ { name: 'London', search_words: [ 'city', 'capital' ], children: [ { name: 'Westminster', search_words: [ 'borough' ], css: { highlight: true } } ] }, { name: 'Manchester', search_words: [ 'city' ] } ] }, { name: 'France', children: [ { name: 'Paris', search_words: [ 'city', 'capital' ] } ] } ] }, 'North America': { children: [ { name: 'Canada', children: [ { name: 'Toronto' }, { name: 'Ottawa' } ] }, { name: 'United States' } ] } } }

console.log(search(myData, mySearchFn('capital')));

// => 2

console.log(myData);

// => { root: { Europe: { children: [ { name: 'Germany' }, { name: 'England', children: [ { name: 'London', search_words: [ 'city', 'capital' ], children: [ { name: 'Westminster', search_words: [ 'borough' ] } ], css: { highlight: true } }, { name: 'Manchester', search_words: [ 'city' ] } ] }, { name: 'France', children: [ { name: 'Paris', search_words: [ 'city', 'capital' ], css: { highlight: true } } ] } ] }, 'North America': { children: [ { name: 'Canada', children: [ { name: 'Toronto' }, { name: 'Ottawa' } ] }, { name: 'United States' } ] } } }

.as-console-wrapper {max-height: 100% !important; top: 0}

使用库对你来说可能不值得，这是一种权衡。如果您对此答案有疑问/想法，请告诉我。

反对回复 2024-01-18

慕尼黑的夜晚无繁华

TA贡献1864条经验获得超6个赞

正如谢谢指出的那样，不一致的数据格式使得为此编写好的代码变得更加困难。我的方法略有不同。我没有转换数据，而是为通用函数编写了一个包装器，以便以更有用的方式处理此输出。

我们从一个函数开始collect，它将递归地处理{name?, search_words?, children?, ...rest}对象，返回与给定谓词匹配的节点并在子节点上重复出现。我们用函数来调用它，search它接受一个搜索词并从中创建一个谓词。（这里我们测试是否name或任何search_term匹配该术语；对于部分匹配、不区分大小写等，这很容易修改。）

然后我们编写我提到的包装器searchLocations。它下降到节点，然后映射并组合调用每个根值.root的结果。search

const collect = (pred) => ({children = [], ...rest}) => [

... (pred (rest) ? [rest] : []),

... children .flatMap (collect (pred))

]

const search = (term) =>

collect (({name = '', search_words = []}) => name == term || search_words .includes (term))

const searchLocations = (locations, term) =>

Object.values (locations .root) .flatMap (search (term))

const locations = {root: {Europe: {children: [{name: "Germany"}, {name: "England", children: [{name: "London", search_words: ["city", "capital"], children: [{name: "Westminster", search_words: ["borough"]}]}, {name: "Manchester", search_words: ["city"]}]}, {name: "France", children: [{name: "Paris", search_words: ["city", "capital"]}]}]}, "North America": {children: [{name: "Canada", children: [{name: "Toronto"}, {name: "Ottawa"}]}, {name: "United States"}]}}}

console .log ('Toronto', searchLocations (locations, 'Toronto'))

console .log ('borough', searchLocations (locations, 'borough'))

console .log ('capital', searchLocations (locations, 'capital'))

.as-console-wrapper {max-height: 100% !important; top: 0}

如果您想要的（听起来像是您可能想要的）与输入的结构相同，仅保留包含匹配项所需的节点，那么我们应该能够从树过滤函数开始执行类似的操作。假期后我会尝试看看。

更新

我确实又看了一遍，希望将树作为一棵树来过滤。代码并没有那么难。但这一次，我确实使用了一个convert函数将您的数据转换为更一致的递归结构。因此，整个对象变成一个数组，根部有两个元素，一个元素为name“欧洲”，另一个元素为name“北美”，每个元素都有其现有children节点。这使得所有进一步的处理变得更加容易。

这里有两个关键函数：

第一个是通用deepFilter函数，它采用谓词和项目数组，这些项目的节点children结构可能类似于其父级，并返回一个新版本，其中包含与谓词及其完整祖先匹配的任何内容。它看起来像这样：

const deepFilter = (pred) => (xs) =>

xs .flatMap (({children = [], ...rest}, _, __, kids = deepFilter (pred) (children)) =>

pred (rest) || kids.length

? [{...rest, ...(kids.length ? {children: kids} : {})}]

: []

)

第二个是专门针对这个问题的：searchLocation。它调用deepFilter使用由搜索项和已讨论的转换结构构造的谓词。它使用convert结构的帮助器，以及search将搜索词转换为谓词的帮助器，该谓词在名称和所有搜索词上查找（不区分大小写）部分匹配。

const searchLocations = (loc, locations = convert(loc)) => (term) =>

term.length ? deepFilter (search (term)) (locations) : locations

这通过用户界面来演示，该用户界面显示嵌套中的位置<UL>，并带有实时过滤位置的搜索框。

例如，如果您在搜索框中输入“w”，您将得到

Europe

England

London (city, capital)

Westminster (borough)

North America

Canada

Ottawa

因为“Westminster”和“Ottawa”是唯一的匹配项。

如果你输入“城市”你会得到

Europe

England

London (city, capital)

Manchester (city)

France

Paris (city, capital)

您可以在此代码片段中看到它的实际效果：

// utility function

const deepFilter = (pred) => (xs) =>

xs .flatMap (({children = [], ...rest}, _, __, kids = deepFilter (pred) (children)) =>

pred (rest) || kids.length

? [{...rest, ...(kids.length ? {children: kids} : {})}]

: []

)

// helper functions

const search = (t = '', term = t.toLowerCase()) => ({name = '', search_words = []}) =>

term.length && (

name .toLowerCase () .includes (term) ||

search_words .some (word => word .toLowerCase() .includes (term))

)

const convert = ({root}) =>

Object.entries (root) .map (([name, v]) => ({name, ...v}))

// main function

const searchLocations = (loc, locations = convert(loc)) => (term) =>

term.length ? deepFilter (search (term)) (locations) : locations

// sample data

// main function specialized to given data

const mySearch = searchLocations(myData)

// UI demo

const format = (locations, searchTerm) => `<ul>${

locations.map(loc => `<li>${

loc.name +

(loc.search_words ? ` (${loc.search_words. join(', ')})` : ``) +

(loc.children ? format(loc.children, searchTerm) : '')

}</li>`)

.join('')

}</ul>`

const render = (locations, searchTerm) =>

document .getElementById ('main') .innerHTML = format (locations, searchTerm)

document .getElementById ('search') .addEventListener (

'keyup',

(e) => render (mySearch (e.target.value))

)

// show demo

render (mySearch (''))

<label>Search: <input type="text" id="search"/></label>

</div>

显然，这并没有使用Vue来生成树，只是进行一些字符串操作和innerHTML. 我把这部分留给你。但它应该显示另一种过滤嵌套结构的方法。

反对回复 2024-01-18

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如何使用 vue/javascript 通过多个属性过滤深度嵌套的 json

如何使用 vue/javascript 通过多个属性过滤深度嵌套的 json

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