根据嵌套列表Java中的日期字段对主列表进行排序

debugcn 发表于 Dev

San San

我在下面列出了对象，并且需要根据日期之前的成分对其进行排序。因此，日期最久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"
            }
        ]
    }
]

我已经尝试过下面的代码，但是它只是对配方中的配料进行排序，而不是列表中的配方对象。Java 8 Streams是否可以实现？

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());

纳曼

由于您需要确保任何成分的历史最悠久bestBefore，因此该食谱应排在列表的最后。您可能需要max在配方的成分中标识（最早的）日期，为此可以创建如下方法：

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

然后对您的配方进行排序，将可以简化操作，例如Comparator<Recipes>为此条件定义a并将其附加到现有比较中。

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());

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