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Accepted
skyho
Asked:2020-12-07 16:46:12 +0000 UTC2020-12-07 16:46:12 +0000 UTC

如何转换 2 个分隔的子字符串 - 在 TreeMap

  • 772

有这样一个文件，几百行，每行必须是TreeMap中的一对。

строка1;сторка2
строка3;сторка5

每行必须匹配 Map 中的一对；

这些行需要从文本文件中读取并传递给 TreeMap 以便它们按顺序排序

    public void read() {

        File dir= new File(path);

       try (
             FileInputStream fileInputStream = new FileInputStream(dir);
             InputStreamReader inputStreamReader = new InputStreamReader(fileInputStream, StandardCharsets.UTF_8);
             BufferedReader reader = new BufferedReader(inputStreamReader)){


           TreeMap <String, String> map = new TreeMap<>();


            while (reader.readLine() != null) {
                System.out.println(reader.readLine());
                // считываем остальные строки в цикле
                String s1 = reader.readLine();

                System.out.println(s1);
            }

        } catch (FileNotFoundException e) {

            logger.error(FILE_NOT_FOUND_ERR, e);

        } catch (IOException e) {

            logger.error(INFO_MESS, e);
        }

        System.out.println("\n" + dir.getAbsolutePath() + "\n");
        logger.info(INFO_MESS);


    }

我尝试了这种方法但没有结果


           Map<String, String> kvs =
                   Arrays.asList(reader.readLine())
                           .stream()
                           .map(elem -> elem.split(";"))
                           .filter(elem -> elem.length==2)
                           .collect(Collectors.toMap(e -> e[0], e -> e[1]));

这就是它的工作原理

    try (
             FileInputStream fileInputStream = new FileInputStream(directory);
             InputStreamReader inputStreamReader = new InputStreamReader(fileInputStream, StandardCharsets.UTF_8);
             BufferedReader reader = new BufferedReader(inputStreamReader)){

           Map <String, String> book = new TreeMap<>();

           String line;

            while (( line = reader.readLine() ) != null) {

                Map <String, String> map =
                        Arrays.asList(line)
                                .stream()
                                .map(elem -> elem.split(";"))
                                .filter(elem -> elem.length==2)
                                .collect(Collectors.toMap(e -> e[0], e -> e[1]));
                System.out.println(map);

                for(Map.Entry<String, String> item : map.entrySet()){
                    book.put(item.getKey(), item.getValue());
                }
            }

           System.out.println(book);

        } catch (FileNotFoundException e) {

            logger.error(FILE_NOT_FOUND_ERR, e);

        } catch (IOException e) {

            logger.error(INFO_MESS, e);
        }

该方法很麻烦，也许有更短的选择？

java

1 个回答

  1. Best Answer

    真的很麻烦......试试这个：

    public class Test {

    private static String FILE_NAME = "test.txt";
    private static String CHARSET_NAME = "windows-1251";

    public static void main(String args[]) throws java.io.IOException {
        Map<String, String> result = Files.lines(Paths.get(FILE_NAME), Charset.forName(CHARSET_NAME))
                .map(l -> l.split(";"))
                .filter(arr->arr.length==2)
                .collect(Collectors.toMap(k -> k[0], v -> v[1], (v1, v2) -> v2, TreeMap::new));
    }

}
    • 2

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