首頁猿問使搜索函數篩選深層嵌套對象項

使搜索函數篩選深層嵌套對象項

JavaScript

倚天杖 2022-08-18 10:09:44

我正在嘗試在我的 React 應用程序中創建一個搜索函數，通過在數組中檢查其中的項目來過濾掉一個項目。我有使用和繪制搜索中特定項目的想法。但是，從后端提取的對象具有用戶將搜索的非常深入嵌套的項目。.filter().includes()下面是一個對象的示例，其中包含我嘗試篩選注釋的項目：const x = [ { owner: "76561198111463361", sold: true, inventory: [ { name: "AK-47 | Redline", //User may look for this price: 1369, }, { name: "StatTrak? AWP | Asiimov", //User may look for this price: 8050, }, ], name: "m", password: "e", }, { owner: "76561198111463361", sold: true, inventory: [ { name: "AK-47 | Redline", //User may look for this price: 1369, }, { name: "StatTrak? AWP | Asiimov", //User may look for this price: 8050, }, ], name: "m", password: "e", },];如何過濾這些？

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3 回答

月關寶盒

TA貢獻1772條經驗獲得超5個贊

不確定要如何過濾它也會共享所需的輸出以獲得更好的清晰度。

var data=[{owner: "76561198111463361",sold: true,inventory: [{ name: "AK-47 | Redline", }, { name: "StatTrak? AWP | Asiimov",}],name: "m",password: "e",},{owner: "76561198111463361",sold: true,inventory: [{ name: "AK-47 | Redline", }, { name: "StatTrak? AWP | Asiimov",}],name: "m",password: "e"}];

var string = 'Redline';

result = data.filter(val=>val.inventory.some(k=>k.name.toUpperCase().includes(string.toUpperCase())));

console.log(result);

如果這解決了您的問題，請告訴我。

反對回復 2022-08-18

手掌心

TA貢獻1942條經驗獲得超3個贊

您可以映射和過濾數據。

const findInventory = (data, fn) => {

return data.reduce((array, item) => {

let inventory = item.inventory;

if (fn !== undefined) {

inventory = inventory.map(fn);

}

array = array.concat(inventory);

return array;

}, []);

};

const data = [{"owner":"76561198111463361","sold":true,"inventory":[{"name":"AK-47 | Redline","price":1369},{"name":"StatTrak? AWP | Asiimov","price":8050}],"name":"m","password":"e"},{"owner":"76561198111463361","sold":true,"inventory":[{"name":"AK-47 | Redline","price":1369},{"name":"StatTrak? AWP | Asiimov","price":8050}],"name":"m","password":"e"}];

// Get objects

console.log(findInventory(data));

// get names

console.log(findInventory(data, (i) => i.name));

第二個例子：

const findInventory = (data, mapper, filter) => {

return data.reduce((array, item) => {

let inventory = item.inventory;

if (mapper !== undefined && filter !== undefined) {

inventory = inventory.map(mapper).filter(filter);

}

array = array.concat(inventory);

return array;

}, []);

};

// Get objects

console.log(findInventory(data));

// get names by filter

console.log(findInventory(data, (i) => i.name, (name) => name.indexOf("AK-") !== -1));

反對回復 2022-08-18

烙印99

TA貢獻1829條經驗獲得超13個贊

你可以用它來過濾掉所有者信息，如果你想要更具體的東西，請評論下來x.filter(item => item.inventory.find(data => data.name.indexOf(searchKeyword) !== -1))

反對回復 2022-08-18

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使搜索函數篩選深層嵌套對象項

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