如何同時導入所有導出常量？

Question

export const getURLPath = (url) => url.split('.com')[1]
export const getSectionName = (path) => path.split('/butik/liste')[1]
export const getIdByName = (sectionName) => sectionMapping.find(item => item.name.toLowerCase() == sectionName).id
export const getNameById = (sectionId) => sectionMapping.find(item => item.id == sectionId).name.toLowerCase()
export const getContentIdFromURL = (path) => path.split('-p-').pop().split('?')[0];
export const getBoutiqueIdFromURL = (path) => path.split('boutiqueId=').pop().split('&')[0];
export const getMerchantIdFromURL = (path) => path.split('merchantId=').pop().split('&')[0];

我想導入右側的所有功能。 我想用他們的純名字。 例如： getURLPath() 就像這樣，沒有前綴等。我怎么做？

Answer 1

 export default { getURLPath : (url) => url.split('.com')[1], getSectionName : (path) => path.split('/butik/liste')[1], getIdByName : (sectionName) => sectionMapping.find(item => item.name.toLowerCase() == sectionName).id, getNameById : (sectionId) => sectionMapping.find(item => item.id == sectionId).name.toLowerCase(), getContentIdFromURL : (path) => path.split('-p-').pop().split('?')[0], getBoutiqueIdFromURL : (path) => path.split('boutiqueId=').pop().split('&')[0], getMerchantIdFromURL : (path) => path.split('merchantId=').pop().split('&')[0], }

Answer 2

如果您不想要utils.getURLPath語法，那么唯一的方法就是將它們全部拼寫出來，即

import {getURLPath, getSectionName, getIdByName, getNameById, /* etc ... */} from '../utils';

Answer 3

你可以使用這樣的東西：

export const fun1...;
export const fun2...;
...

// This is not even mandatory I think but leaves u the choice to use prefix at some point. As u wish
export default {
  fun1,
  fun2,
}

然后將它們導入到您的其他文件中，例如：

import {
  fun1,
  fun2,
  ...
} from '/your/file';

// Then use them
fun1();