如何解析Java Collections.sort（）比较方法违反其常规约定异常

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;
        });

---

关于这个问题，我浏览了谷歌的网页，上面说我在比较大的物体和小的物体。我尝试反转object的顺序，但没有成功。

如果元素中的任何一个未能解析，则只能捕获一次异常

考虑以下时间的三次装运：
A-12:34
B-34:56
C-null

使用比较器，

compare（A，C）

将返回0，

compare（B，C）

将返回0，但是

compare（A，B）

将返回非零结果，因此违反了传递性的一般约定

一种简单的方法是使用

比较器。比较
语法以分别解析每个元素：

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;
                   })))
              );

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的是哪个StringUtils，
com.sun.deploy.util.StringUtils
没有
.isEmpty（）
方法。所以我重新实现了它

以下是解决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());
}

输出：

[发货{shipmentPickingTime='12:00'}，发货{shipmentPickingTime='12:03'}，发货{shipmentPickingTime='12:02'}]
[Shippings{shipmentPickingTime='12:00'}，Shippings{shipmentPickingTime='12:02'}，Shippings{shipmentPickingTime='12:03'}]
能否请共享装运对象的字段和类型？您的装运确实有一个时间戳，上面写着“null”（不是
null
）？比较器必须符合此处列出的要求：@GhostCat是的，它包含空值，但我已使用检查来防止空值。我还能做什么？@JollyRoger该字段是字符串类型，我正在将其转换为日期进行排序。如果时间为空，这会给我空指针异常。我不想抛出任何异常，只想按ASC顺序对列表进行排序。@ShreyaMahajan对此表示抱歉
nullsLast
总是让我绊倒。请参阅我编辑的答案，了解正确的使用方法。
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());
}

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 + '\'' +
                '}';
    }
}