使用Java8流过滤列表

Question

我有一个客户列表对象如下。

import java.util.Arrays;
import java.util.List;
import java.util.stream.Collectors;

public class CheckForResource {

    public static void main(String[] args) {
        Customer john = new Customer("111", "P_1", "daily",
                "create", "table_1", "03-05-2020",
                "03-05-2020", "140");

        Customer mary = new Customer("111", "P_1", "daily",
                "delete", "table_1", "03-05-2020",
                "03-05-2020", "30");

        Customer joseph = new Customer("222", "P_2", "weekly",
                "create", "table_2", "03-05-2020",
                "03-05-2020", "50");

        Customer jason = new Customer("222", "P_2", "daily",
                "update", "table_2", "03-05-2020",
                "03-05-2020", "40");

        Customer mario = new Customer("111", "P_1", "weekly",
                "create", "table_1", "03-05-2020",
                "03-05-2020", "20");

        Customer danny = new Customer("111", "P_1", "monthly",
                "update", "table_1", "03-05-2020",
                "03-05-2020", "100");

        List<CheckForResource.Customer> customers = Arrays.asList(john, mary, joseph, jason, mario, danny);

    }


    public static class Customer {

        final String Id;
        final String pCode;
        final String usageType;
        final String operation;
        final String resource; 
        final String startTime;
        final String endTime;
        final String value;

        public Customer(String id, String pCode, String usageType, String operation,
                        String resource, String startTime, String endTime, String value) {
            Id = id;
            this.pCode = pCode;
            this.usageType = usageType;
            this.operation = operation;
            this.resource = resource;
            this.startTime = startTime;
            this.endTime = endTime;
            this.value = value;
        }
    }
}

如果列表中的每一个子句中至少有1个条目，我想返回true。

1. customerId="111", operation="create", usageType="daily"
2. customerId="111", operation="create", usageType="monthly"
3. customerId="111", operation="删除", usageType="每日"
4. customerId="111", operation="删除", usageType="月度"
5. customerId="111", operation="update", usageType="daily"
6. customerId="111", operation="update", usageType="monthly"

我如何使用steams来实现这个功能？

Answer 1

你可以使用 .anyMatch()

Predicate<Customer> p = c -> c.getId().equals("111") &&
                        (c.getUsageType().equals("daily") || c.getUsageType().equals("monthly")) &&
                        (c.getOperation().equals("create") || c.getOperation().equals("delete") || c.getOperation().equals("update"));
boolean result = customers.stream().filter(p)
                          .map(c -> c.getId().concat(c.getOperation()).concat(c.getUsageType()))  // Map to create hash string for detect unique
                          .distinct().count() == 6;  // distinct and check the count.

Answer 2

试试这样。

• 过滤掉id==111之后，你有六种不同的可能性。
• 因此，如果你构建一个谓词列表，你可以做以下的事情
• 如果你有六个条目，那么它就满足了条件。

创建一个基于谓词列表的 operation 和 usageType

List<Predicate<String>> preds = toPredicateList(
        new String[] { "create", "delete", "update" },
        new String[] { "daily", "monthly"});

确定是否满足条件。

boolean result = 
        customers
            .stream()
            .filter(c -> c.Id.equals("111"))
            .map(c -> c.operation + c.usageType)
            .filter(preds.stream().reduce(k -> false,
                                Predicate::or))
            .collect(Collectors.toSet()).size() == 6;

System.out.println(result);

构建谓词列表

public List<Predicate<String>> toPredicateList(String[] ops,
            String[] types) {

        List<Predicate<String>> list = new ArrayList<>();
        for (String o : ops) {
            for (String t : types) {
                list.add(s -> s.equals(o + t));
            }
        }
    }
    return list;
}