0

假设您有来自关系数据库的数据postscategories

// category_id on categories.id
const posts = [
  { id: '1', title: 'dream pc', category_id: '1' },
  { id: '2', title: 'my dream house design', category_id: '1' },
  { id: '3', title: 'bing chillin', category_id: '2' },
]

const categories = [
  { id: '1', name: 'wishlist' }, 
  { id: '2', name: 'favorites' }
]

这是我的路线格式:/categories/[category]/[page]

目录结构为pages/categories/[category]/[page].js.

所以它会产生这些路径:

  • /categories/wishlist/1
  • /categories/wishlist/2
  • /categories/favorites/1

我已经尝试过了,但paths返回了一个空数组:

// pages/categories/[category]/[page].js

export async function getStaticPaths() {
    const paths = []

    // I'm using supabase
    const { data: categories, error } = await supabase.from('categories').select('id, name')

    categories.forEach(async (c) => {
        const { data, error, count } = await supabase
            .from('posts')
            .select('id', { count: 'exact', head: true })
            .eq('category_id', c.id)

        for (let i = 0; i < count; i++) {
            // This should push the path params to the paths variable, but it didn't
            paths.push({
                params: {
                    // For simplicity, 1 post per page
                    page: i + 1,
                    categoryId: c.id,
                    category: c.name
                }
            })
        }
    })

    // It returns an empty array []
    console.log(paths)

    return { paths, fallback: false }
}
4

1 回答 1

1

异步调用不适用于Array.forEach,而是我使用for (category of categories)

工作代码:

// pages/categories/[category]/[page].js

export async function getStaticPaths() {
    const paths = []

    // I'm using supabase
    const { data: categories, error } = await supabase.from('categories').select('id, name')

    // Use for/of instead of forEach
    for (category of categories) {
        const { data, error, count } = await supabase
            .from('posts')
            .select('id', { count: 'exact', head: true })
            .eq('category_id', c.id)

        for (let i = 0; i < count; i++) {
            paths.push({
                params: {
                    page: (i + 1).toString(),
                    categoryId: category.id,
                    category: category.name
                }
            })
        }
    })

    return { paths, fallback: false }
}
于 2021-10-14T14:09:46.290 回答