Question

例如

Maze0.bmp (0,0) (319,239) 65 120
Maze0.bmp (0,0) (319,239) 65 120 (254,243,90)  
Maze0.bmp (0,0) (319,239) 65 120 (254,243,90) (0,0,0)
Maze0.bmp (0,0) (319,239) 65 120 (254,243,90) (0,0,0) (11,33,44)

我想获得maze0.bmp和所有数字。我有：

Pattern pattern = Pattern.compile("([A-z][^\\s]*)\\s+\\((\\d+),(\\d+)\\)\\s+\\((\\d+),(\\d+)\\)\\s+(\\d+)\\s+(\\d+)\\s+(\\((\\d+),(\\d+),(\\d+)\\)\\s*)"); 
BufferedReader stdin = new BufferedReader(new InputStreamReader( System.in));
String input;
Matcher matcher = null; 
boolean isMatched = false; 
while (!isMatched) {
    System.out.println("Please enter right format\n");
    input = stdin.readLine(); 
    matcher = pattern.matcher(input); 
    while(matcher.find()) {
        isMatched = true; 
        for (int i = 1; i <= matcher.groupCount(); ++i) 
            System.out.println(matcher.group(i));
    }
}

但这是正确的。例如，如果我的输入是

Maze0.bmp (0,0) (319,239) 65 120 (254,243,90) (0,0,0)

它无法获取最后一个元组（0,0,0）。

Answer 1

这是我能想到的最好的。请注意，我使用了两种模式，因为出于某种原因，Java拒绝捕获重复组（如果有人碰巧知道原因，请发表评论）。

final Pattern outerPattern = Pattern.compile("(.*?) \\((\\d+),(\\d+)\\) \\((\\d+),(\\d+)\\) (\\d+) (\\d+)(.*)");
final Pattern optionalTouplePattern = Pattern.compile(" \\((\\d+),(\\d+),(\\d+)\\)");

final BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
boolean isMatched;

do
{
    System.out.println("Please enter right format:");
    Matcher m = outerPattern.matcher(stdin.readLine());
    if (isMatched = m.find())
    {
        System.out.println(String.format("name='%s', first touple: [%s,%s], second touple: [%s,%s], first single number: %s, second single number: %s", m.group(1), m.group(2), m.group(3), m.group(4), m.group(5), m.group(6), m.group(7)));
        m = optionalTouplePattern.matcher(m.group(8));
        while(m.find())
        {
            System.out.println(String.format("+ optional touple: [%s,%s,%s]", m.group(1), m.group(2), m.group(3)));
        }
    }
}while(!isMatched);

Answer 2

好的，对不起，我得修改一下。 java匹配器似乎不喜欢它在正则表达式的编译时无法确定的模式计数。但这有效（经过测试）：

Matcher m = Pattern.compile("\\((\\d+),(\\d+),(\\d+)\\)").matcher("(23,56,78) (54,22,11)");
while(m.find())
{
  for(int i = 1; i <= m.groupCount(); ++i)
  System.out.println(m.group(i));
}

Answer 3

我不知道java中匹配的上下文，但我非常了解正则表达式。试试这个背景：

while matching BITMAP records is not done
  ("
     ([A-z][^\s])            'maze.bmp'   ~ group 1
     \s+
     \(  (\d+),(\d+)  \)     '0' '0'      ~ group 2,3
     \s+
     \( (\d+),(\d+) \)       '319'  '239' ~ group 4,5
     \s+
     (\d+)                   '65'         ~ group 6
     \s+
     (\d+)                   '120'        ~ group 7
     \s+
     (
       (?: \( \d+,\d+,\d+ \) \s+ )+       '(254,243,90) (0,0,0) '     ~ group 8
     )
  ") - context = global
{
      // save to BITMAP.array (groups 1 - 7)
      copy group 8 to variable '(254,243,90) (0,0,0) '
      new matching of TUPLES, group 8 is the regex subject for this new match
         ("
            (\d+)
         ") - context = global

      append TUPLES.array (254 243 90 0 0 0)
      to BITMAP.array (maze.bmp 0 0 319 239 65 120 <append> 254 243 90 0 0 0)

     // do next BITMAP record
}

如何用这种格式编写一个java模式：任何字符（int，int）（int，int）数字的任意数量（int，int，int）

3 个答案: