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)
我想要获取 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)
无法获取最后一个元组(0,0,0)。
最佳答案
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);
关于java - 如何以这种格式编写java模式 : any characters (int, int) (int,int) number number any number of (int,int,int),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4911671/