列表包含另一个列表，主列表应根据逻辑进行修改

Question

我有一个对象列表，如下所示。如果任何成分超过了使用截止日期，则生成的列表不应包含食谱。这可以通过Java Streams实现吗。

我试着像下面这样做，但遇到了一个编译错误。非常感谢你的回复，因为我一直在努力寻找答案。

List<Recipes> filteredList =
recipeList.stream().filter(recipes -> recipes.getIngredients().stream().filter(ingredients -> ingredients.getUseBy().before(lunchDate)).collect(Collectors.toList()));

[
{
"id": 1,
"recipeName": "Pasta",
"createdDate": "2020-11-22T00:00:00.000+00:00",
"ingredients": [
{
"ingredientId": 1,
"ingredientName": "Shells",
"useBy": "2020-11-20",
"bestBefore": "2020-11-18"
},
{
"ingredientId": 2,
"ingredientName": "Mozaralla Cheese",
"useBy": "2020-12-20",
"bestBefore": "2020-11-23"
}
]
},
{
"id": 2,
"recipeName": "Burger",
"createdDate": "2020-11-22T00:00:00.000+00:00",
"ingredients": [
{
"ingredientId": 3,
"ingredientName": "Burger Bun",
"useBy": "2020-12-20",
"bestBefore": "2020-12-23"
},
{
"ingredientId": 4,
"ingredientName": "Beef Pattie",
"useBy": "2020-12-20",
"bestBefore": "2020-12-23"
},
{
"ingredientId": 5,
"ingredientName": "Tomato",
"useBy": "2020-12-20",
"bestBefore": "2020-12-23"
}
]
},
{
"id": 3,
"recipeName": "Chicken Salad",
"createdDate": "2020-11-22T00:00:00.000+00:00",
"ingredients": [
{
"ingredientId": 6,
"ingredientName": "Salad Mix",
"useBy": "2020-12-20",
"bestBefore": "2020-11-15"
},
{
"ingredientId": 7,
"ingredientName": "Chicken",
"useBy": "2020-12-20",
"bestBefore": "2020-12-23"
}
]
}
]

Answer 1

Stream().filter()将Predicate<T>作为参数，并返回由该流中与给定谓词匹配的元素组成的流。您的要求可以改写为：

List<Recipes> filteredList = recipeList.stream().filter(predicate: recipes does not contain any ingredient past useBy date).collect(Collectors.toList()));

您的问题被更改为编写匹配的谓词：配方不包含任何超过useBy日期的成分。(对于任何配方，计算通过useBy日期的成分，计数应为0。(

你可以使用另一个stream().filter(predicate: count ingredient after useBy date, and it should be 0)

recipes -> recipes.getIngredients().stream()
.filter(ingredients -> ingredients.getUseBy().after(lunchDate))
.collect(Collectors.toList()).size() == 0

然后将谓词添加到第一个筛选器表达式中。最后的表达式是这样的：

List<Recipes> filteredList = recipeList.stream().filter(
recipes -> recipes.getIngredients().stream()
.filter(ingredients -> ingredients.getUseBy().after(lunchDate))
.collect(Collectors.toList()).size() == 0
).collect(Collectors.toList()));

Answer 2

这里的工作示例(不使用JSON(。

考虑以下内容：

List<Recipe> filteredList = 
recipes.stream()
.filter(r -> ! r.getIngredients()
.stream()
.anyMatch(i -> i.getUseBy().before(lunchDate)))                         
.collect(Collectors.toList());

每个配方的描述是：

• 获取配料
• 在lunchDate之前用useBy查找任何"坏"成分
• 如果有任何"坏"成分，则排除(通过!运算符(配方