这是因为的贪婪，
*随之而来的是回溯。
字串：
This order was placed for QT3000! OK?
正则表达式：
(.*)(\d+)(.*)
我们都知道那
.*是贪婪的，并且尽可能匹配所有字符。因此，第一个
.*匹配所有字符，直到最后一个匹配
?，然后按顺序回溯以提供匹配。正则表达式中的下一个模式是
d+，因此它回溯到一个数字。一旦找到数字，就
d+匹配该数字，因为这里满足条件（
d+匹配一个或多个数字）。现在，第一个
(.*)捕获
This order was placed forQT300，随后的
(\d+)捕获捕获
0位于
!符号之前的数字。
现在，下一个模式将
(.*)捕获剩余的所有字符
!<space>OK?。
m.group(1)指的是出现在组索引1内的字符，并
m.group(2)指代索引2，就像它继续一样。
在这里查看演示。
获得所需的输出。
String line = "This order was placed for QT3000! OK?"; String pattern = "(.*)(\d{2})(.*)"; // Create a Pattern object Pattern r = Pattern.compile(pattern); // Now create matcher object. Matcher m = r.matcher(line); while(m.find( )) { System.out.println("Found value: " + m.group(1)); System.out.println("Found value: " + m.group(2)); System.out.println("Found value: " + m.group(3)); }输出：
Found value: This order was placed for QT30Found value: 00Found value: ! OK?
(.*)(\d{2})，按顺序回溯最多两位数以提供匹配。改变你的模式，
String pattern = "(.*?)(\d+)(.*)";
要获得类似的输出，
Found value: This order was placed for QTFound value: 3000Found value: ! OK?
?在
*部队
*进行非贪婪的比赛之后。
使用额外的捕获组从单个程序获取输出。
String line = "This order was placed for QT3000! OK?";String pattern = "((.*?)(\d{2}))(?:(\d{2})(.*))";Pattern r = Pattern.compile(pattern); Matcher m = r.matcher(line); while(m.find( )) { System.out.println("Found value: " + m.group(1)); System.out.println("Found value: " + m.group(4)); System.out.println("Found value: " + m.group(5)); System.out.println("Found value: " + m.group(2)); System.out.println("Found value: " + m.group(3) + m.group(4)); System.out.println("Found value: " + m.group(5)); }输出：
Found value: This order was placed for QT30Found value: 00Found value: ! OK?Found value: This order was placed for QTFound value: 3000Found value: ! OK?
