Question

我对Retrofit 2和Android / Backend编程相对较新，我想知道如何正确地将我的JSON模式转换为Java学校项目的Java。

所以情况是我在我的应用程序中需要一些“挑战”。目前的结构有点复杂，我甚至不确定我们是否正常工作，但我会在这里试一试。

这些是我的REST后端的mongoose架构：

挑战：所以这个对象包含所有正常属性和对challengeType aswel的引用，因为我们的挑战是由多态性定义的。

var ChallengeSchema = new Schema({
_id: ObjectId,
name: String,
description: [String],
challengeType:{
    type: Schema.ObjectId, ref: 'ChallengeType'
},
state: Number,
difficulty: Number
});

现在我们的challengeTypes被定义为：

var ChallengeTypeSchema = new Schema({
    _id: ObjectId
}),
  RecipeChallengeSchema = new Schema({
    _id: ObjectId,
    joined: Number,
    dietTypeMaker: {
        type: Schema.ObjectId, ref: 'DietTypeMaker'
    }
}),
  FastfoodChallengeSchema = new Schema({
    _id: ObjectId,
    joined: Number
}),
  ThursdayVeggieDayChallengeSchema = new Schema({
    _id: ObjectId,
    joined: Number,
    dietTypeMaker1: {
        type: Schema.ObjectId, ref: 'DietTypeMaker'
    },
    dietTypeMaker2: {
        type: Schema.ObjectId, ref: 'DietTypeMaker'
    }
}),
  RestaurantChallengeSchema = new Schema({
    _id: ObjectId,
    joined: Number,
    city: {
        type: Schema.ObjectId, ref: 'City'
    },
    restaurantTypeMaker: {
        type: Schema.ObjectId, ref: 'RestaurantTypeMaker'
    }
});
module.exports = mongoose.model('ChallengeType',ChallengeTypeSchema,'challengeTypes');
module.exports = mongoose.model('RecipeChallenge', RecipeChallengeSchema, 'challengeTypes', ['ChallengeType']);
module.exports = mongoose.model('FastfoodChallenge', FastfoodChallengeSchema, 'challengeTypes', ['RecipeChallenge']);
module.exports = mongoose.model('ThursdayVeggieDayChallenge', ThursdayVeggieDayChallengeSchema, 'challengeTypes', ['RecipeChallenge']);
module.exports = mongoose.model('RestaurantChallenge', RestaurantChallengeSchema, 'challengeTypes', ['ChallengeType']);

这些与DietTypeMaker或RestaurantTypeMaker相关联，根据活跃的挑战确定我们的食谱或餐馆。

var RestaurantTypeMakerSchema = new Schema({
_id: ObjectId,
eatingPlace: {
    restaurants: [
        {
            type: Schema.ObjectId, ref: 'Restaurant'
        }
    ]
},
gastronomic: {
    restaurants: [
        {
            type: Schema.ObjectId, ref: 'Restaurant'
        }
    ]
},
snackbar: {
    restaurants: [
        {
            type: Schema.ObjectId, ref: 'Restaurant'
        }
    ]
},
catering: {
    restaurants: [
        {
            type: Schema.ObjectId, ref: 'Restaurant'
        }
    ]
}
});

和

var DietTypeMakerSchema = new Schema({
_id: ObjectId,
vegetarian: {
    recipe: {
        type: Schema.ObjectId, ref: 'Recipe'
    }
},
glutenFree: {
    recipe: {
        type: Schema.ObjectId, ref: 'Recipe'
    }
},
sugarFree: {
    recipe: {
        type: Schema.ObjectId, ref: 'Recipe'
    }
},
childFriendly: {
    recipe: {
        type: Schema.ObjectId, ref: 'Recipe'
    }
}
});

每个都指食谱或餐馆：

var RecipeSchema = new Schema({
_id: ObjectId,
title: String,
difficultyRate: Number,
ingredients: [
    {
        name: String,
        qty: String
    }
],
description: [String],
extraInfo: {
    preparationTime: Number,
    cookDifficulty: Number,
    recipeType: Number,
    dietTypes: [Number],
    isFastFood: Boolean
},
image: {
    data: Buffer, contentType: String
}
});

和

var RestaurantSchema = new Schema({
_id: ObjectId,
name: String,
streetName: String,
houseNumber: Number,
city: {
    type: Schema.ObjectId, ref: 'City'
},
restaurantTypes: [Number],
preferences: [Number],
websiteLink: String,
emailaddress: String,
description: [String]
});

这一切都需要转换为我们的应用程序的Java代码。现在我知道如何制作pojos，我已经为食谱和餐厅制作了它们。我可以使用Retrofit 2从我的后端正确地获取它们。现在我坚持的是我们对所有模式引用的复杂结构...

我们是在制造复杂的东西还是甚至可能...... ???

使用Retrofit 2的JSON嵌套模式引用

0 个答案: