如何解析Java集合。sort（）比较方法违反了它的一般约定异常

Question

我正在使用 Collections.sort 根据时间字段按升序对列表进行排序。下面是代码

private String getShipmentInpickingTime(List<Shipments> shipments) {
        logger.info("in getShipmentInpickingTime");
        DateFormat sdf = new SimpleDateFormat("hh:mm");

        Collections.sort(shipments, (o1, o2) -> {
            try {
                if ((!"null".equals(o1.getShipmentinpickingtime())
                        && !StringUtils.isEmpty(o1.getShipmentinpickingtime()))
                        && (!"null".equals(o2.getShipmentinpickingtime())
                                && !StringUtils.isEmpty(o2.getShipmentinpickingtime()))) {
                    return sdf.parse(o1.getShipmentinpickingtime()).compareTo(sdf.parse(o2.getShipmentinpickingtime()));
                }
            } catch (ParseException e) {
                e.printStackTrace();
            }
            int count1 = 0;
            return count1;
        });

此方法引发异常--

java.lang.IllegalArgumentException: Comparison method violates its general contract.

关于这个问题，我浏览了谷歌，上面说我在比较大对象和小对象。我试着颠倒物体的顺序，但没有成功。

Answer 1

您没有提到哪行代码会引发此错误。我不确定您在这里使用了哪些StringUtils，com.sun.deploy.util。StringUtils没有。isEmpty（）方法。所以我重新实现了它。

以下是针对您的问题的工作代码:

public static void main(String[] args){
    getShipmentInpickingTime(new ArrayList<Shipments>(){{
        add(new Shipments("12:00"));
        add(new Shipments("12:03"));
        add(new Shipments("12:02"));
    }});
}

private static void getShipmentInpickingTime(List<Shipments> shipments) {
    DateFormat sdf = new SimpleDateFormat("hh:mm");

    System.out.println(shipments.toString());
    Collections.sort(shipments, (o1, o2) -> {
        try {
            if ((!"null".equals(o1.getShipmentPickingTime())
                    && !(o1.getShipmentPickingTime() == null || o1.getShipmentPickingTime().length() < 1))
                    && (!"null".equals(o2.getShipmentPickingTime())
                    && !(o2.getShipmentPickingTime() == null || o2.getShipmentPickingTime().length() < 1))) {
                return sdf.parse(o1.getShipmentPickingTime()).compareTo(sdf.parse(o2.getShipmentPickingTime()));
            }
        } catch (ParseException e) {
            e.printStackTrace();
        }
        return 0;
    });

    System.out.println(shipments.toString());
}

货件类内容：

public class Shipments {

    private String shipmentPickingTime;

    public Shipments(String shipmentPickingTime) {
        this.shipmentPickingTime = shipmentPickingTime;
    }

    public String getShipmentPickingTime() {
        return shipmentPickingTime;
    }

    public void setShipmentPickingTime(String shipmentPickingTime) {
        this.shipmentPickingTime = shipmentPickingTime;
    }

    @Override
    public String toString() {
        return "Shipments{" +
                "shipmentPickingTime='" + shipmentPickingTime + '\'' +
                '}';
    }
}

输出:

[装运{装运拣件时间='12：00'}， 装运{装运拣件时间='12：03'}， 装运{装运拣货时间='12：02'}] [装运{装运拣件时间='12：00'}， 装运{装运拣件时间='12：02'}， 装运{装运拣件时间='12：03'}]

Answer 2

如果任一元素解析失败，则仅捕获一次异常。

考虑三次装运，时间如下:< br> A - 12:34
B - 34:56
C -空< br >

使用您的比较器，比较（A， C）将返回0，比较（B， C）将返回0，但比较（A， B）将返回非零结果，从而违反传递性的一般契约。

一种简单的方法是使用Comparator.comparing语法分别解析每个元素：

DateFormat sdf = new SimpleDateFormat("hh:mm");

shipments.sort(Comparator.comparing(
                   Shipments::getShipmentinpickingtime, 
                   Comparator.nullsLast(Comparator.comparing(time -> {
                       try {
                           if (!"null".equals(time) && !StringUtils.isEmpty(time)) {
                               return sdf.parse(time);
                           }
                       } catch (ParseException ignoe) {
                           // Not a valid time
                       }
                       return null;
                   })))
              );