Java，正则表达式HasNext以空行开头，支持多平台 [英] Java, Regular Expression HasNext starts with empty line, multi-platform support

问题描述

我需要在Unix和Windows上处理以下文件:

I need to process the following file on Unix and Windows:

a;b
c;d;e;f;g
c;d;e;f;g
c;d;e;f;g

a;b
c;d;e;f;g
c;d;e;f;g
c;d;e;f;g

a;b

a;b
c;d;e;f;g
c;d;e;f;g
c;d;e;f;g

我需要处理下面包含一个数据块的a;b. 例如第三 a;b不应该被处理.

i need to process a;b that contain a block of data underneath. e.g. the third a;b shouldn't be processed.

当前，我正在使用Java扫描程序通过以下正则表达式来分隔文件中此类文本:

currently i am delimiting by using the following regular expression this type of text in a file using Java scanner:

Scanner fileScanner = new Scanner(file);
        try{

            fileScanner.useDelimiter(Pattern.compile("^$", Pattern.MULTILINE));

            while(fileScanner.hasNext()){
                String line;
                while ((line = fileScanner.nextLine()).isEmpty());
                InputStream is = new ByteArrayInputStream(fileScanner.next().getBytes("UTF-8"));
...

这仍将为 1/3 a;b委派ByteArrayInputStream的空输入.

This will still delegate for the third a;b the empty input into the ByteArrayInputStream.

我可以检查fileScanner.next()的第一行是否为空行，然后执行nextLine()语句和后续的continue语句吗?

Hoe may i check if the first line of fileScanner.next() is an empty line and then execute nextLine() statement and a following a continue statement?

推荐答案

使用正则表达式模式

(?m)^(?:.+(?:\\r?\\n|\\Z)){2,}

匹配两个或多个非空行，或者换句话说，两个或多个(?:...){2,}行包含一个或多个字符.+，后跟新行\\r?\\n或(?:...|...)字符串\\Z的末尾.

which matches two or more non-empty lines, or other words two or more (?:...){2,} lines that contain one or more characters .+ followed by new line \\r?\\n or (?:...|...) end of string \\Z.

多行修饰符(?m)表示^匹配每行的开头，而不仅仅是字符串的开头.

Multiline modifier (?m) means that ^ matches a beginning of each line, not just the beginning of the string.

String str = "...";

Pattern p = Pattern.compile("(?m)^(?:.+(?:\\r?\\n|\\Z)){2,}");
Matcher m = p.matcher(str);
while (m.find()) {
  String match = m.group();
  System.out.println(match);
}

请参见 此演示 .

See this demo.

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