参考http://www.cnblogs.com/deerchao/archive/2006/08/24/zhengzhe30fengzhongjiaocheng.html
- package sunfa.lx;
- import java.util.Arrays;
- import java.util.regex.Pattern;
- /**
- * java.util.regex包负责对字符序列进行正则表达式匹配
- * Pattern负责编译
- * Matcher负责匹配
- * ---------------
- * 正则表达式:
- *1、元字符 好比char,紧紧代表一个字符;
- *2、某些字符,比如\、.、*等这样的字符在正则表达式中已经被作为标记使用了,如果你想作为非元字符使用的话就得进行转义,转义的方法是在这些字符前面加\,
- * 比如\就变成\\,.就成为\.
- *3、重复的元字符有:*,+,?,{n},{n,},{n,m},这个标记是用来修饰它前面的那个字符的。
- *4、字符类,或称为分组[],比如[0-9],[0-9A-Za-z]用-表示范围
- */
- public class PatternDemo1 {
- public static void main(String[] args) {
- /**1、元字符的[]*/
- //[]属于元字符系列,所谓元字符简单的说就是字符,就是一个char,像bc就不是一个char了
- // method2(new String[]{"abt","act","adt","abct","bt"}, "a[bcd]t");
- //[]的区间,-的两头代表区间,多区间这样写,比如[0-9a-zA-Z]
- // method2(new String[]{"a1t","a3t","a01t","abt","2t"}, "a[0-9]t");
- //补集用^表示,在二元运算符中这是非的意思,尽管用了^,at还是不能通过,所以元字符[]必须有且仅有一个字符
- // method2(new String[]{"a0t","a2t","abt","at"}, "a[^0246]t");
- /**2、元字符的? + * */
- //?的目标是它前面的一个字符,元字符?表示它前面的一个字符出现0次或1次
- // method2(new String[]{"","a","aa","at","a1t"}, "a?");
- //元字符+的目标是它前面的一个字符,它表示它前面的那个字符出现1次或多次
- // method2(new String[]{"","a","aa","aaaaa","at","a1t"}, "a+");
- //元字符*的目标是它前面的一个字符,它表示它前面那个字符出现0次或多次,*可以说是?和+的并集
- // method2(new String[]{"","a","aa","aaaaa","at","a1t"}, "a*");
- /** .元字符能匹配任何字符(换行\r除外),所以用.*可以匹配换行外的任何字符串 */
- // method2(new String[]{"","a","aa","aaaaa","at","a1t","\t","\r","n"}, ".");
- // method2(new String[]{"","a","aa","aaaaa","at","a1t","\t","\r","n"}, ".*");
- /** 数量{n},{n,},{n,m}*/
- //{n}表示它前面的字符重复n次,并且只重复n次,重复n-1次或n+1次都是不可以的
- // method2(new String[]{"","a","aa","aaaaa","at","a1t"}, "a{2}");
- //{n,}是对{n}的扩展,表示重复n次(包含n次)以上
- // method2(new String[]{"","a","aa","aaaaa","at","a1t"}, "a{2,}");
- //{n,m}表示数目在n到m范围内,包含两头
- // method2(new String[]{"","a","aa","aaa","aaaa","aaaaa","at"}, "a{2,4}");
- // method2(new String[]{"12345-1234","12345"}, "\\d{5}-\\d{4}|\\d{5}");
- //
- // method2(new String[]{"12345-123","12345"}, "\\d{5}|\\d{5}-\\d{4}");
- // method2(new String[]{"211","12345"}, "2[0-4]\\d");
- // method2(new String[]{"1","12345"}, "[01]?\\d\\d?");
- method2(new String[]{"1","12345","12"}, "[0-9]");
- }
- private static void method2(String[] a,String regex){
- Pattern p = Pattern.compile(regex);
- for (int i = 0; i < a.length; i++) {
- System.out.println(a[i]+","+p.matcher(a[i]).matches());
- }
- }
- private static void method1(){
- String str = "2011-11-12";
- String pat = "\\d{4}-\\d{2}-\\d{2}";
- System.out.println(Pattern.compile(pat).matcher(str).matches());
- String str1 = "a1b22c333d4444e55555f";
- //按数字来分割
- String[] str1Arr = Pattern.compile("\\d+").split(str1);
- System.out.println(Arrays.toString(str1Arr));
- //数字全部替换成_
- System.out.println(Pattern.compile("\\d+").matcher(str1).replaceAll("_"));
- //去除所有的空格
- System.out.println("\\s去除所有的空格:"+Pattern.compile("\\s").matcher("aa b c d ").replaceAll(""));
- System.out.println("5-10个字符:"+Pattern.compile(".{5,10}").matcher("12345").matches());
- System.out.println("多匹配:"+Pattern.compile("\\bhi.*Lucy\\b").matcher("hi后面不远处跟着一个Lucy").matches());
- System.out.println("多匹配2:"+Pattern.compile("^\\d{5,12}$").matcher("12345678a").matches());
- //直接调用String类提供的方法更加的方便
- System.out.println(Arrays.toString(str1.split("\\d+")));
- System.out.println("2011-11-12".matches("\\d{4}-\\d{2}-\\d{2}"));
- }
- }
java正则表达式及java.util.regex包的学习
时间: 2024-10-17 18:30:16