使用集合排序时出现问题

时间:2012-08-21 11:21:19

标签: java sorting collections compare

我正在尝试从列表中对数据进行排序,该列表的字段ID为:

field id :field_38637
field id :field_38469
field id :field_38468
field id :field_38638
field id :field_38637
field id :field_38469
field id :field_38468
field id :field_38468
field id :field_38469
field id :field_38468
field id :field_38637

列表数据的大小是11,对于我正在进行排序的上面列表数据如下:

    Collections.sort(copedppList, new Comparator() {

        @Override
        public int compare(Object o1, Object o2) {
            Field oo1 = (Field ) o1;
            Field oo2 = (Field ) o2;
        if(oo1.getFieldId()!=oo2.getFieldId()){}    
                    return 1;
                      else 
                         return 0;


    });


   for (int i = 0; i < copedppList.size(); i++) {

            Field pp = (Field) copedppList
                    .get(i);
            System.out.println(pp.getFieldID());
        }

排序后我得到了列表

field_38468
field_38468
field_38468
field_38468
field_38468
field_38469
field_38469
field_38469
field_38637
field_38637
field_38637

编辑: 类似的字段ID在排序后被组合在一起,但它应按照以38637,38469,38468开头的输入列表进行分组,这里有什么问题?我想保持传入列表的顺序。

2 个答案:

答案 0 :(得分:7)

您没有在比较器中进行排序,只检查相等性。这可能会更好:

public int compare(Object o1, Object o2) {
    Field oo1 = (Field ) o1;
    Field oo2 = (Field ) o2;

    return oo1.getFieldID().compareTo(oo2.getFieldID());
}

(假设getFieldID是非空字符串)

ps:您似乎没有使用泛型,它可以让您从代码中删除所有这些强制转换。

修改

您实际上想要对项目进行分组,而不是对它们进行排序。最简单的方法是使用2个列表(未测试):

public static void main(String[] args) throws InterruptedException {
    List<Field> originalList = new ArrayList<Field>();
    originalList.add(new Field("field_38637"));
    originalList.add(new Field("field_38469"));
    originalList.add(new Field("field_38468"));
    originalList.add(new Field("field_38638"));
    originalList.add(new Field("field_38637"));
    originalList.add(new Field("field_38469"));
    originalList.add(new Field("field_38468"));
    originalList.add(new Field("field_38468"));
    originalList.add(new Field("field_38469"));
    originalList.add(new Field("field_38468"));
    originalList.add(new Field("field_38637"));

    System.out.println(originalList);
    List<Field> groupedList = new ArrayList<Field>(originalList.size());
    while (!originalList.isEmpty()) {
        String id = originalList.get(0).getFieldID();
        for (Iterator<Field> it = originalList.iterator(); it.hasNext();) {
            Field f = it.next();
            if (f.getFieldID().equals(id)) {
                it.remove();
                groupedList.add(f);
            }
        }
    }

    System.out.println(groupedList);
}

public static class Field {

    private final String fieldID;

    public Field(String fieldID) {
        this.fieldID = fieldID;
    }

    public String getFieldID() {
        return fieldID;
    }

    @Override
    public String toString() {
        return "Field{" + "fieldID=" + fieldID + '}';
    }
}        

答案 1 :(得分:2)

阅读compare()的文档。您应该返回-1,0或1,具体取决于元素ab之间的关系。您只返回0或1,因此排序将不起作用。

尝试阅读http://www.javapractices.com/topic/TopicAction.do?Id=10http://docs.oracle.com/javase/7/docs/api/java/util/Comparator.html

尝试

public int compare(Object o1, Object o2) 
{
  Field oo1 = (Field ) o1;
  Field oo2 = (Field ) o2;
  return oo1.getFieldID().compareTo(oo2.getFieldID());
}