对馆藏进行排序

Question

我正在尝试根据前两列的值对文件内容进行排序：

我的文件内容是：

档案名称：ComplexMapFileReader.txt

内容是：

B 12 hgjhgjgjgjgjgjhgjgjgjgjgjgj
B 3 jgjdSAGdjgsdKJADJgjfgAJFL
B 6 jygtjgjgjgjgjgjhgjhbj
C 7 uiyuiyikykk
C 2 wrteyytyuiiiiyjhg
A 1 hikhkhkhkjhkhkhjkhkh
A 4 khkhkkhkhkhkhkh
A 2 khkhkhkhkhkhkhkhkhkhkhkhh
A 11 jhgjsGJDGjAGFjgfjgjfgj
A 5 gjgjgsadgajgjDGkkhkhdDKjhkjd
B 5 hdskdhkljdhKAJHDKjahsd
A 12 kjhkdjhfknkjflkjfsdlkjf
C 3 kjhfekJEHFKHefkh
B 34 khkhkjhkjhkqlwwjljlllljj
A 21 iuhEWHRIekhrlkHRKErhwkrhk;h;kfnk.nfas
C 11 jlklkqwwlklklklllklkkkkjjj         

我的代码是：

package CollectionExamples;

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.HashMap;
import java.util.Map;
import java.util.Map.Entry;
import java.util.TreeMap;

public class MapFileReader {

    private static final String FILENAME = "D:\\ComplexMapFileReader.txt";

    public static void main(String[] args) {

        BufferedReader br = null;
        FileReader fr = null;

        Map<Integer, String> innerMap = null;
        Map<String, TreeMap> outerMap = new TreeMap();
        try {
            fr = new FileReader(FILENAME);
            br = new BufferedReader(fr);
            String sCurrentLine;
            br = new BufferedReader(new FileReader(FILENAME));

            try {
                while ((sCurrentLine = br.readLine()) != null) {

                    String[] s = sCurrentLine.split(" ");

                    if (!outerMap.containsKey(s[0])) {
                        innerMap = new TreeMap();
                        innerMap.put(Integer.valueOf(s[1]), s[2]);
                        outerMap.put(s[0], (TreeMap) innerMap);
                    } else {
                        innerMap.put(Integer.valueOf(s[1]), s[2]);
                        outerMap.put(s[0], (TreeMap) innerMap);
                    }

                }

                String s0 = null;
                String s1 = null;
                String s2 = null;

                for (Entry<String, TreeMap> map1 : outerMap.entrySet()) {

                    s0 = map1.getKey();

                    for (Entry<Integer, String> map2 : innerMap.entrySet()) {

                        s1 = " "+map2.getKey() + "";
                        s2 =" "+ map2.getValue();
                        System.out.println("Each line equals:" + s0 + s1 + s2);
                        System.out.println("#############");

                    }

                }

            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

    }

}

上面的代码没有给我想要的结果。 但是当上面的文件以某种有序的方式排列时（字母将所有B和A在一起），则文件内容如下所示：

B 12 hgjhgjgjgjgjgjhgjgjgjgjgjgj
B 3 jgjdSAGdjgsdKJADJgjfgAJFL
B 6 jygtjgjgjgjgjgjhgjhbj
C 7 uiyuiyikykk
C 2 wrteyytyuiiiiyjhg
A 1 hikhkhkhkjhkhkhjkhkh
A 4 khkhkkhkhkhkhkh
A 2 khkhkhkhkhkhkhkhkhkhkhkhh
A 11 jhgjsGJDGjAGFjgfjgjfgj
A 5 gjgjgsadgajgjDGkkhkhdDKjhkjd

结果是所需的结果（按字母顺序，然后按数字顺序）：

输出：

Each line equals:A 1 hikhkhkhkjhkhkhjkhkh
#############
Each line equals:A 2 khkhkhkhkhkhkhkhkhkhkhkhh
#############
Each line equals:A 4 khkhkkhkhkhkhkh
#############
Each line equals:A 5 gjgjgsadgajgjDGkkhkhdDKjhkjd
#############
Each line equals:A 11 jhgjsGJDGjAGFjgfjgjfgj
#############
Each line equals:B 1 hikhkhkhkjhkhkhjkhkh
#############
Each line equals:B 2 khkhkhkhkhkhkhkhkhkhkhkhh
#############
Each line equals:B 4 khkhkkhkhkhkhkh
#############
Each line equals:B 5 gjgjgsadgajgjDGkkhkhdDKjhkjd
#############
Each line equals:B 11 jhgjsGJDGjAGFjgfjgjfgj
#############
Each line equals:C 1 hikhkhkhkjhkhkhjkhkh
#############
Each line equals:C 2 khkhkhkhkhkhkhkhkhkhkhkhh
#############
Each line equals:C 4 khkhkkhkhkhkhkh
#############
Each line equals:C 5 gjgjgsadgajgjDGkkhkhdDKjhkjd
#############
Each line equals:C 11 jhgjsGJDGjAGFjgfjgjfgj
#############

但是当文件的内容如下时，我没有得到想要的结果。 我知道是地图，在以下情况下它将覆盖先前的值。 但是有什么更好的方法可以解决这个问题。

B 12 hgjhgjgjgjgjgjhgjgjgjgjgjgj
    B 3 jgjdSAGdjgsdKJADJgjfgAJFL
    B 6 jygtjgjgjgjgjgjhgjhbj
    C 7 uiyuiyikykk
    C 2 wrteyytyuiiiiyjhg
    A 1 hikhkhkhkjhkhkhjkhkh
    A 4 khkhkkhkhkhkhkh
    A 2 khkhkhkhkhkhkhkhkhkhkhkhh
    A 11 jhgjsGJDGjAGFjgfjgjfgj
    A 5 gjgjgsadgajgjDGkkhkhdDKjhkjd
    B 5 hdskdhkljdhKAJHDKjahsd
    A 12 kjhkdjhfknkjflkjfsdlkjf
    C 3 kjhfekJEHFKHefkh
    B 34 khkhkjhkjhkqlwwjljlllljj
    A 21 iuhEWHRIekhrlkHRKErhwkrhk;h;kfnk.nfas
    C 11 jlklkqwwlklklklllklkkkkjjj         

Answer 1

我认为这是用Record对象表示每一行的更好方法。 实际上，我们不需要TreeMap进行排序。 如果您想添加更多字段或更改比较逻辑，它将更加方便。 对于记录列表，我们只需要实现Comparable接口即可实现您的比较逻辑。

例如：

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class MapFileReader {

    private static final String FILENAME = "D:\\ComplexMapFileReader.txt";


    private static class Record {
        private char ch;
        private int num;
        private String string;

        public void setCh(char ch) {
            this.ch = ch;
        }

        public char getCh() {
            return ch;
        }

        public void setNum(int num) {
            this.num = num;
        }

        public int getNum() {
            return num;
        }

        public String getString() {
            return string;
        }

        public void setString(String string) {
            this.string = string;
        }

    }

    public static void main(String[] args) {

        BufferedReader br;
        try {
            String sCurrentLine;
            br = new BufferedReader(new FileReader(FILENAME));

            List<Record> records = new ArrayList<>();

            try {
                while ((sCurrentLine = br.readLine()) != null) {

                    String[] s = sCurrentLine.split(" ");

                    if (s.length != 3) {
                        // Filter invalid records
                        continue;
                    }

                    Record r = new Record();
                    r.ch = s[0].charAt(0);
                    r.num = Integer.parseInt(s[1]);
                    r.string = s[2];

                    records.add(r);
                }

                Collections.sort(records, new Comparator<Record>() {
                    @Override
                    public int compare(Record o1, Record o2) {
                        if (o1.ch == o2.ch) {
                            return o1.num < o2.num ? -1 : (o1.num == o2.num ? 0 : 1);
                        }
                        return o1.ch < o2.ch ? -1 : 1;
                    }
                });

                for (Record record : records) {
                    System.out.println(String.format("Each line equals: %s %s %s", record.ch, record.num, record.string));
                    System.out.println("#############");
                }

            } catch (IOException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }

        } catch (FileNotFoundException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

    }

}