我有一個這樣的物件：
{
"idMeal": "52795",
"strMeal": "Chicken Handi",
"strDrinkAlternate": null,
"strCategory": "Chicken",
"strArea": "Indian"
}
我正在嘗試使用“idMeal”的值獲取其他專案（即 strMeal 和 strCategory）的值。我怎樣才能做到這一點？
uj5u.com熱心網友回復：
據我了解，您要過濾食物物件串列，只需將它們放入一個陣列中并像這樣過濾：
[obj1, obj2, obj3].filter(o => o.idMeal="52795")
uj5u.com熱心網友回復：
首先，您應該說明您嘗試過的內容。然后，您可以使用一個簡單的條件：
let cat;
let area;
if (obj.idMeal === myValue) {
cat = obj.strCategory;
area = obj.strArea;
}
有更多的方法可以做到這一點。看看.map()，.filter()但那些是用于陣列的。
uj5u.com熱心網友回復：
實作這一點的最簡單方法是根據鍵值對過濾資料，然后將請求的欄位縮減為一個新物件。
const mealData = [{
"idMeal": "52795",
"strMeal": "Chicken Handi",
"strDrinkAlternate": null,
"strCategory": "Chicken",
"strArea": "Indian",
}];
const filterAndMap = (data, key, value, fields) =>
data
.filter((record) => record[key] === value)
.map(record => fields.reduce((acc, field) =>
({ ...acc, [field]: record[field] }), {}));
const result = filterAndMap(mealData, 'idMeal', '52795', ['strMeal', 'strCategory']);
console.log(result);或者，您可以傳遞過濾器函式和映射器函式：
const mealData = [{
"idMeal": "52795",
"strMeal": "Chicken Handi",
"strDrinkAlternate": null,
"strCategory": "Chicken",
"strArea": "Indian",
}];
const filterAndMap = (data, filterFn, mapperFn) =>
data.filter(filterFn).map(mapperFn);
const result = filterAndMap(
mealData,
({ idMeal }) => idMeal === '52795',
({ strMeal, strCategory }) => ({ strMeal, strCategory }));
console.log(result);在上面的例子中，我們可以洗掉filterAndMap函式filter并map直接鏈接。
顯示代碼片段
const mealData = [{
"idMeal": "52795",
"strMeal": "Chicken Handi",
"strDrinkAlternate": null,
"strCategory": "Chicken",
"strArea": "Indian",
}];
const result = mealData
.filter(({ idMeal }) => idMeal === '52795')
.map(({ strMeal, strCategory }) => ({ strMeal, strCategory }));
console.log(result);轉載請註明出處，本文鏈接：https://www.uj5u.com/ruanti/359695.html
標籤：javascript json 映射 过滤