嵌套列表Java中基于日期字段的主列表排序

Question

我有下面的物品清单，需要根据日期前的成分进行分类。因此，最古老的、最好的、最早的食谱应该排在最后

 [
    {
        "id": 2,
        "recipeName": "Burger",
        "createdDate": "2020-11-22T00:00:00.000+00:00",
        "ingredients": [
            {
                "ingredientId": 3,
                "ingredientName": "Burger Bun",
                "category": "",
                "useBy": "2020-12-20",
                "bestBefore": "2020-12-23"
            },
            {
                "ingredientId": 4,
                "ingredientName": "Beef Pattie",
                "category": "",
                "useBy": "2020-12-20",
                "bestBefore": "2020-12-23"
            },
            {
                "ingredientId": 5,
                "ingredientName": "Tomato",
                "category": "",
                "useBy": "2020-12-20",
                "bestBefore": "2020-09-10"
            }
        ]
    },
    {
        "id": 3,
        "recipeName": "Chicken Salad",
        "createdDate": "2020-11-22T00:00:00.000+00:00",
        "ingredients": [
            {
                "ingredientId": 7,
                "ingredientName": "Chicken",
                "category": "",
                "useBy": "2020-12-20",
                "bestBefore": "2020-12-23"
            },
            {
                "ingredientId": 6,
                "ingredientName": "Salad Mix",
                "category": "",
                "useBy": "2020-12-20",
                "bestBefore": "2020-11-15"
            }
        ]
    }
]

我尝试了下面的代码，但它只是对配方中的成分进行排序，而不是列表中的配方对象。这在Java8流中是可能的吗

List<Recipes> finalList = filteredList.stream().sorted((o1, o2) -> (int) (o1.getId() - o2.getId()))
        .map(recipes -> {
            List<Ingredients> in = recipes.getIngredients().stream()
                    .sorted(Comparator.comparing(Ingredients::getBestBefore).reversed()).collect(Collectors.toList());
            recipes.setIngredients(in);
            return recipes;
        }).collect(Collectors.toList());

Answer 1

因为你需要确保如果任何成分有最早的bestBefore日期，那么该配方应该排在列表的最后您需要在配方的成分中确定max（最早的）日期，为此您可以创建如下方法：

private static Date findOldestBestBeforeForRecipe(Recipes recipes) {
    return recipes.getIngredients()
            .stream()
            .map(Ingredients::getBestBefore)
            .max(Comparator.naturalOrder())
            .orElseThrow(() -> new IllegalArgumentException("recipes without ingredients!!"));
}

然后对你的食谱进行排序，这会简化，比如为这个标准定义一个Comparator<Recipes>，并将它们附加到现有的比较中

Comparator<Recipes> ingredientsBestBeforeDate = (o1, o2) ->
        findOldestBestBeforeForRecipe(o2).compareTo(findOldestBestBeforeForRecipe(o1));

List<Recipes> finalList = filteredList.stream()
        .sorted(Comparator.comparingInt(Recipes::getId)
                .thenComparing(ingredientsBestBeforeDate))
        .collect(Collectors.toList());