Java将文件读入ArrayList？

Question

你如何将文件的内容读入ArrayList<String>Java中？

这是文件内容:

cat
house
dog
.
.
.

只需阅读每个单词ArrayList.

Answer 1

这个java代码读入每个单词并将其放入ArrayList:

Scanner s = new Scanner(new File("filepath"));
ArrayList<String> list = new ArrayList<String>();
while (s.hasNext()){
    list.add(s.next());
}
s.close();

使用s.hasNextLine()并且s.nextLine()如果您想逐行而不是逐字阅读.

Answer 2

与commons-io的单行:

List<String> lines = FileUtils.readLines(new File("/path/to/file.txt"), "utf-8");

与番石榴一样:

List<String> lines = 
     Files.readLines(new File("/path/to/file.txt"), Charset.forName("utf-8"));

Answer 3

您可以使用:

List<String> list = Files.readAllLines(new File("input.txt").toPath(), Charset.defaultCharset() );

来源:Java API 7.0

Answer 4

我发现的最简单的形式是......

List<String> lines = Files.readAllLines(Paths.get("/path/to/file.txt"));

Answer 5

在Java 8中,您可以使用流和Files.lines:

List<String> list = null;
try (Stream<String> lines = Files.lines(myPathToTheFile))) {
    list = lines.collect(Collectors.toList());
} catch (IOException e) {
    LOGGER.error("Failed to load file.", e);
}

或者作为一个函数,包括从文件系统加载文件:

private List<String> loadFile() {
    List<String> list = null;
    URI uri = null;

    try {
        uri = ClassLoader.getSystemResource("example.txt").toURI();
    } catch (URISyntaxException e) {
        LOGGER.error("Failed to load file.", e);
    }

    try (Stream<String> lines = Files.lines(Paths.get(uri))) {
        list = lines.collect(Collectors.toList());
    } catch (IOException e) {
        LOGGER.error("Failed to load file.", e);
    }
    return list;
}

Answer 6

List<String> words = new ArrayList<String>();
BufferedReader reader = new BufferedReader(new FileReader("words.txt"));
String line;
while ((line = reader.readLine()) != null) {
    words.add(line);
}
reader.close();