java.lang.IllegalStateException: Duplicate key-编程知识

java.lang.IllegalStateException: Duplicate key

序言

最近监控扫描出我们项目的某些异常信息，报错java.lang.IllegalStateException: Duplicate key xxx，看到异常来自stream流，然后定位看了一下是某位同事的代码使用stream流把List转Map集合出现重复的key异常信息。List集合A对象来源于某个接口的返回，使用A对象的uuid成员变量作为key，理论上uuid作为唯一标识不应该有重复。
所以正确的做法是：
1）找该接口对应责任人，定位看List对象A的uuid为什么出现重复；
2）查看本项目代码中的异常来源；

java.util.stream.Collectors

Java 8版本引入Stream流式数据处理方式，使得可以对集合进行更简单高效的操作。
API文档链接如下：

oracle.com/javase/8/doc/stream
microsoft.com/api/java.util.stream.collectors

异常根因

首先看一段示例代码，

import lombok.AllArgsConstructor;
import lombok.Builder;
import lombok.Data;
import lombok.NoArgsConstructor;import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collector;
import java.util.stream.Collectors;public class DuplicateKeyDemo {@AllArgsConstructor@NoArgsConstructor@Data@Builderprivate static class Employee {private int identifierId;private String name;private int age;}public static void main(String[] args) {List<Employee> employList = new ArrayList<>();employList.add(new Employee(1, "Mike", 25));employList.add(new Employee(2, "Mary", 26));employList.add(new Employee(3, "Jack", 28));employList.add(new Employee(4, "Tom", 23));employList.add(new Employee(5, "Lucy", 21));employList.add(new Employee(6, "Jim", 26));employList.add(new Employee(7, "David", 29));employList.add(new Employee(8, "Jack", 22));employList.add(new Employee(8, "Jack", 25));Collector<Employee, ?, Map<String, Integer>> collector = Collectors.toMap(Employee::getName, Employee::getAge);Map<String, Integer> nameAgeIgnoreRepeatMap = employList.stream().collect(collector);System.out.println(nameAgeIgnoreRepeatMap);}
}

抛出的异常信息如下，

Exception in thread "main" java.lang.IllegalStateException: Duplicate key 28at java.util.stream.Collectors.lambda$throwingMerger$0(Collectors.java:133)at java.util.HashMap.merge(HashMap.java:1254)at java.util.stream.Collectors.lambda$toMap$58(Collectors.java:1320)at java.util.stream.ReduceOps$3ReducingSink.accept(ReduceOps.java:169)at java.util.ArrayList$ArrayListSpliterator.forEachRemaining(ArrayList.java:1382)at java.util.stream.AbstractPipeline.copyInto(AbstractPipeline.java:482)at java.util.stream.AbstractPipeline.wrapAndCopyInto(AbstractPipeline.java:472)at java.util.stream.ReduceOps$ReduceOp.evaluateSequential(ReduceOps.java:708)at java.util.stream.AbstractPipeline.evaluate(AbstractPipeline.java:234)at java.util.stream.ReferencePipeline.collect(ReferencePipeline.java:499)at com.hust.zhang.stream.DuplicateKeyDemo.main(DuplicateKeyDemo.java:43)

直接点到HashMap的merge方法时，Map中已经存在28岁的Jack，新来的一条数据是22岁的Jack，那么在执行remappingFunction.apply(old.value, value)时就会报错java.lang.IllegalStateException: Duplicate key 28
在这里插入图片描述

解决方案

如果假设我们在这个Map里就是可以被覆盖，那么应该怎么做？

import lombok.AllArgsConstructor;
import lombok.Builder;
import lombok.Data;
import lombok.NoArgsConstructor;import java.util.ArrayList;
import java.util.List;
import java.util.Map;
import java.util.stream.Collector;
import java.util.stream.Collectors;public class DuplicateKeyDemo {@AllArgsConstructor@NoArgsConstructor@Data@Builderprivate static class Employee {private int identifierId;private String name;private int age;}public static void main(String[] args) {List<Employee> employList = new ArrayList<>();employList.add(new Employee(1, "Mike", 25));employList.add(new Employee(2, "Mary", 26));employList.add(new Employee(3, "Jack", 28));employList.add(new Employee(4, "Tom", 23));employList.add(new Employee(5, "Lucy", 21));employList.add(new Employee(6, "Jim", 26));employList.add(new Employee(7, "David", 29));employList.add(new Employee(8, "Jack", 22));employList.add(new Employee(8, "Jack", 25));Collector<Employee, ?, Map<String, Integer>> antiCollisionCollector = Collectors.toMap(Employee::getName,Employee::getAge, (oldValue, newValue) -> oldValue);Map<String, Integer> nameAgeMap = employList.stream().collect(antiCollisionCollector);System.out.println(nameAgeMap);}
}

针对原来代码改了一行，在Collectors.toMap方法后面追加(oldValue, newValue) -> oldValue，表示当出现冲突时取初始出现值，如下图，执行remappingFunction.apply(old.value, value)方法时返回初始值28，
在这里插入图片描述
如果改成(oldValue, newValue) -> newValue则在本示例中返回重复对象的末尾值25。