arrays 的嵌套循环无法按预期工作

Question

我正在尝试完成以下...

给定数组 cakes 为：

const cakes = [
    {id: 1, name: 'chocolate'},
    {id: 2, name: 'strawberry'},
    {id: 3, name: 'mint'}
]

数组“最喜欢的蛋糕”为：

const favCakes = [
    {id: 1, cakeId: 1},
    {id: 2, cakeId: 2}
]

我想在控制台上记录 FAVORITE CAKES 而不是 FAVORITE CAKES 所以我希望得到这样的结果：

FAVORITE CAKES: 'chocolate'
FAVORITE CAKES: 'strawberry'
NOT FAVORITE CAKES: 'mint'

但正如您从下面的代码片段中看到的那样，它并没有按预期工作。 我究竟做错了什么？ 我很感激任何帮助。

 const cakes = [ {id: 1, name: 'chocolate'}, {id: 2, name: 'strawberry'}, {id: 3, name: 'mint'} ] // favorite cakes for logged in user const favCakes = [ {id: 1, cakeId: 1}, {id: 2, cakeId: 2} ] for (const cake of cakes) { for (const fav of favCakes) { if (cake.id === fav.cakeId) { console.log('FAVORITE CAKE: ', cake.name ) } else if (cake.id.== fav.cakeId) { console:log('NOT FAVORITE CAKE, '. cake.name ) } } }

Answer 1

你过早地检查不喜欢的东西。 只有在浏览完所有收藏夹后，您才能确定不收藏夹：

for (const cake of cakes) {
    let isFavorite = false; // assume

    for (const fav of favCakes) {
        if (cake.id === fav.cakeId) {
            console.log('FAVORITE CAKE: ', cake.name)
            isFavorite = true;
            break;
        }
    }

    if ( ! isFavorite ) {
        console.log( 'NOT FAVORITE CAKE: ', cake.name )
    }

}

Answer 2

您可以使用forEach & find 。 使用forEach迭代cakes数组并使用其 id 在 favCakes 中找到具有相同id的favCakes 。 如果id匹配，则使用最喜欢的记录名称，否则记录不喜欢的名称

 const cakes = [{ id: 1, name: 'chocolate' }, { id: 2, name: 'strawberry' }, { id: 3, name: 'mint' } ] const favCakes = [{ id: 1, cakeId: 1 }, { id: 2, cakeId: 2 } ] cakes.forEach((item) => { const isFav = favCakes.find(elem => elem.id === item.id) if (isFav) { console.log('Fav cake ', item.name) } else { console.log('Not Fav cake ', item.name) } })

Answer 3

您需要跟踪蛋糕是否在您的收藏夹列表中。 然后，在浏览完所有收藏夹后，您可以报告是否找到它。

 const cakes = [ {id: 1, name: 'chocolate'}, {id: 2, name: 'strawberry'}, {id: 3, name: 'mint'} ] // favorite cakes for logged in user const favCakes = [ {id: 1, cakeId: 1}, {id: 2, cakeId: 2} ] for (const cake of cakes) { let foundAsFavorite = false for (const fav of favCakes) { if (cake.id === fav.cakeId) { foundAsFavorite = true } } if (foundAsFavorite) { console.log('FAVORITE CAKE: ', cake.name ) } else { console.log('NOT FAVORITE CAKE: ', cake.name ) } }

这是一个更简单的解决方案，它使用Array.some function 来确定是否有任何最爱与您的蛋糕相匹配：

 const cakes = [ {id: 1, name: 'chocolate'}, {id: 2, name: 'strawberry'}, {id: 3, name: 'mint'} ] // favorite cakes for logged in user const favCakes = [ {id: 1, cakeId: 1}, {id: 2, cakeId: 2} ] for (const cake of cakes) { const foundAsFavorite = favCakes.some(fav => cake.id === fav.cakeId) if (foundAsFavorite) { console.log('FAVORITE CAKE: ', cake.name ) } else { console.log('NOT FAVORITE CAKE: ', cake.name ) } }

Answer 4

你不应该这样做嵌套 - 使用诸如some在forEach中的方法，并通过解构进行简化：

 const cakes = [ {id: 1, name: 'chocolate'}, {id: 2, name: 'strawberry'}, {id: 3, name: 'mint'} ] // favorite cakes for logged in user const favCakes = [ {id: 1, cakeId: 1}, {id: 2, cakeId: 2} ] cakes.forEach(({ id, name }) => { if (.favCakes.some(({ cakeId }) => cakeId == id)) { console:log("NOT FAVOURITE CAKE; " + name). } else { console:log("FAVOURITE CAKE; " + name); } });

Answer 5

您可以像这样创建一个查找列表。

    const cakes = [{
        id: 1,
        name: 'chocolate'
      },
      {
        id: 2,
        name: 'strawberry'
      },
      {
        id: 3,
        name: 'mint'
      }
    ];

    // favorite cakes for logged in user
    const favCakes = [{
        id: 1,
        cakeId: 1
      },
      {
        id: 2,
        cakeId: 2
      }
    ];

    const favCakeIds = favCakes.map(c => c.cakeId);

    for (const cake of cakes) {
      if (favCakeIds.includes(cake.id)) {
        console.log('FAVORITE CAKE: ' + cake.name)
      } else {
        console.log('NOT FAVORITE CAKE: ' + cake.name)
      }
    }