猫鼬发现所有然后找到所有其他模式每个

Question

在我的应用我需要找到分类的话，我想这个分类图书

const CategorySchema = mongoose.Schema({ 
    displayname: String, 
    category: String 
}); 

module.exports = mongoose.model('Category', CategorySchema); 

} 

const BookSchema = mongoose.Schema({ 
    name: String, 
    img: String, 
    category: { 
     type: mongoose.Schema.Types.ObjectId, 
     ref: 'Category' 
    } 
}); 

module.exports = mongoose.model('Books', BookSchema); 

我需要这样的

categories = [ 
{ 
    category: 'News', 
    books: [{name:'',id''},{}..] 
}, 
{ 
    category: 'Sports', 
    books: [{name:'',id''},{}..] 
} 
]; 

响应所以我的代码看起来像这样

Router.get('/categories/', (req, res) => { 
    Category.find({}, (err, categories) => { 
     Promise.all(categories.map(category => { 
      return Book.find({category: category._id}).then(books => { 
       return { 
        category: category, 
        books: books 
       }; 
      }) 
     })).then(categories => { 
      res.json(categories); 
     }) 
    }) 
}); 

请有人知道如果有更好的解决方案，这个查询？ 谢谢大家！

Answer 1

从http://mongoosejs.com/docs/populate.html

Book. 
    find({}). 
    populate('category'). 
    exec(function (err, story) { 
    if (err) return handleError(err); 
    // Do other things 
    }); 

Answer 2

这个怎么样？

Router.get('/categories/', (req, res) => { 
    Book.find().populate('category').then(books => { 
    var categories = {} 
    for(var i = 0; i < books.length; i++){ 
     categories[books[i].category.displayname] = { 
     category: books[i].category.displayname 
     books: categories[books[i].category.displayname].books ? categories[books[i].category.displayname].books.push(books[i]) : [books[i]] 
     }; 
    } 
    var result = []; 
    for(var key in categories){ 
     result.push(categories[key]) 
    } 
    return result; 
    }); 
}); 

这可能是更漂亮，但它只会让一个请求到你的数据库，并应工作你想