Java对象集合按照指定元素顺序排序

SmileNicky

发布于 2024-12-23 11:00:05

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文章被收录于专栏：Nicky's blogNicky's blog

需求背景

最近在对一个集合列表的数据进行排序，需求是要集合数据按照一个排序状态值进行排序，而这个状态值，不是按照从小到大这样的顺序排序的，而是要按照特定的顺序，比如按照1, 0, 2的顺序排的，所以需要自定义一个排序工具类来实现

解决方案

写一个排序工具类

package cn.nzp.ems.ecb.server.business.screen;


import java.io.Serializable;
import java.util.Comparator;
import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.Stream;

public class CustomizeNumComparator implements Comparator<Integer>, Serializable {

    private static final long serialVersionUID = 6076418974912880405L;

    List<Integer> numList = Stream.of(1, 0, 2).collect(Collectors.toList());


    @Override
    public int compare(Integer o1, Integer o2) {
        int i1 = numList.indexOf(o1);
        int i2 = numList.indexOf(o2);
        return (i1 == -1 || i2 == -1) ? (i2 - i1) : i1 - i2;
    }
}

测试一下排序

@Test
public void testCompare() {
	List<RecordDto> list = Lists.newArrayList();
	RecordDto record01= new RecordDto();
	record01.setOrderNum(2);
	record01.setSName("测试名称1");

	RecordDto record02= new RecordDto();
	record02.setOrderNum(0);
	record02.setSName("测试名称2");

	RecordDto record03 = new RecordDto();
	record03.setOrderNum(1);
	record03.setSName("测试名称3");

	RecordDto record04= new RecordDto();
	record04.setOrderNum(2);
	record04.setSName("测试名称4");

	RecordDto record05= new RecordDto ();
	record05.setOrderNum(0);
	record05.setSName("测试名称5");

	list.add(record01);
	list.add(record02);
	list.add(record03);
	list.add(record04);
	list.add(record05);


	list = list.stream()
			.sorted(Comparator.comparing(RecordDto::getOrderNum, new CustomizeNumComparator()))
			.collect(Collectors.toList());
	list.stream().forEach(e -> {
		System.out.println(e.getName() +", " +
				e.getOrderNum());
	});
}

跑一下测试结果

如果不想写一个排序工具类，也可以创建一个Comparator即可

public void testCompare() {
	List<Integer> numList = Stream.of(1, 0, 2).collect(Collectors.toList());
	Comparator<RecordDto> comparingIndex = Comparator.comparing(record -> {
		final int index = numList.indexOf(record.getOrderNum());
		return index == -1 ? Long.MAX_VALUE : index;
	});
	list = list.stream()
			.sorted(comparingIndex.thenComparing(RecordDto::getCreateTime, Comparator.nullsLast(Date::compareTo)))
			.collect(Collectors.toList());
}

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原始发表：2024-12-23，如有侵权请联系 cloudcommunity@tencent.com 删除

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