如何从 System.in / System.console() 构建 Java 8 流？
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/29154405/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
How to build a Java 8 stream from System.in / System.console()?
提问by Georgy Ivanov
Given a file, we can transform it into a stream of strings using, e.g.,
给定一个文件，我们可以使用例如，将其转换为字符串流
Stream<String> lines = Files.lines(Paths.get("input.txt"))
Can we build a stream of lines from the standard input in a similar way?
我们可以以类似的方式从标准输入构建一个行流吗？
回答by Georgy Ivanov
A compilation of kocko's answer and Holger's comment:
kocko 的回答和 Holger 的评论的汇编：
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
Stream<String> stream = in.lines().limit(numberOfLinesToBeRead);
回答by Adrian
you can use just Scannerin combination with Stream::generate:
您可以仅Scanner与Stream::generate以下组合使用：
Scanner in = new Scanner(System.in);
List<String> input = Stream.generate(in::next)
.limit(numberOfLinesToBeRead)
.collect(Collectors.toList());
or (to avoid NoSuchElementExceptionif user terminates before limit is reached):
或（避免NoSuchElementException用户在达到限制之前终止）：
Iterable<String> it = () -> new Scanner(System.in);
List<String> input = StreamSupport.stream(it.spliterator(), false)
.limit(numberOfLinesToBeRead)
.collect(Collectors.toList());
回答by Konstantin Yovkov
Usually the standard input is read line by line, so what you can do is store all the read line into a collection, and then create a Streamthat operates on it.
通常标准输入是逐行读取的，所以你可以做的就是将所有读取的行存储到一个集合中，然后创建一个对其Stream进行操作的集合。
For example:
例如：
List<String> allReadLines = new ArrayList<String>();
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String s;
while ((s = in.readLine()) != null && s.length() != 0) {
allReadLines.add(s);
}
Stream<String> stream = allReadLines.stream();
