跨多个阵列查找数据的更好方法

Question

getComponentById: (state) => (componentId) => {
return state.articles
.filter(article => Object.keys(article).some(key => {
return ['maps', 'charts', 'tables'].includes(key);
}))
.reduce((acc, article) => {
acc = article.components?.find(c => c.id == componentId);
if (acc) return acc;
acc = article.maps?.find(c => c.id == componentId);
if (acc) return acc;
acc = article.charts?.find(c => c.id == componentId);
if (acc) return acc;
acc = article.tables?.find(c => c.id == componentId);
if (acc) return acc;
})
}

不知道是否有更好的方法来重写它，因为组件列表可能会增加，所以不断添加行感觉是错误的。

Answer 1

如果id是唯一的，您可以查看每篇文章中的每一个键吗？

如果我对你的数据结构的猜测很接近，你应该能够做一些类似的事情

let articles = [
{
maps: [{ id: 1, name: 'map1' }, { id: 2, name: 'map2' }],
charts: [{ id: 3, name: 'charts1' }, { id: 4, name: 'charts2' }],
tables: [{ id: 5, name: 'tables1' }, { id: 6, name: 'tables2' }]
},
{
maps: [{ id: 7, name: 'map3' }, { id: 8, name: 'map4' }],
charts: [{ id: 9, name: 'charts3' }, { id: 10, name: 'charts4' }],
tables: [{ id: 11, name: 'tables3' }, { id: 12, name: 'tables4' }]
}
]

let getComponentById = (componentId) => {
let result = null;  

articles.forEach(article => {
Object.keys(article).forEach(key => {
let component = article[key].find(x=> x.id == componentId);
if(component) {
result = component;
}
});
});

return result;
}
console.log(getComponentById(3));
console.log(getComponentById(12));

Answer 2

感谢@IrKenInvader的回答，我从他那里复制了数据。

我使用for loop是因为一旦找到组件，就可以提前返回，而无需检查其余数据。

let state = {
articles: [
{
maps: [
{ id: 1, name: "map1" },
{ id: 2, name: "map2" },
],
charts: [
{ id: 3, name: "charts1" },
{ id: 4, name: "charts2" },
],
tables: [
{ id: 5, name: "tables1" },
{ id: 6, name: "tables2" },
],
},
{
maps: [
{ id: 7, name: "map3" },
{ id: 8, name: "map4" },
],
charts: [
{ id: 9, name: "charts3" },
{ id: 10, name: "charts4" },
],
tables: [
{ id: 11, name: "tables3" },
{ id: 12, name: "tables4" },
],
},
],
};
const getComponentById = state => componentId => {
for (let i = 0; i < state.articles.length; i++) {
const filteredKey = Object.keys(state.articles[i]).filter(key =>
["maps", "charts", "tables"].includes(key)
);
for (let j = 0; j < filteredKey.length; j++) {
const foundComponent = state.articles[i][filteredKey[j]].find(
a => a.id == componentId
);
if (foundComponent) return foundComponent;
}
}
return null;
};
const output = getComponentById(state)(12);
console.log(output);