Question

java 怎么提取 字符串中 匹配一条正则表达式的字符串

java 怎么提取 字符串中 匹配一条正则表达式的字符串

Answer 1

String str = "*******************";
Pattern pattern = Pattern.compile("\\{@[A-Za-z0-9]+#\\}");
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
System.out.println(matcher.group());
}

Answer 2

如果是要查找一条“正则表达式”的字符串，你需要把查找的关键字，转义。
如要查找“.*”
你需要写成“\.\*”

Answer 3

public static List<LabelValueBean> getList(String str) {
List<String> list = new ArrayList<String>();
Pattern pattern = Pattern.compile("\\{@[A-Za-z0-9]+#\\}");
if(UtilValidate.isEmpty(str)){
return null;
}
Matcher matcher = pattern.matcher(str);
while (matcher.find()) {
if (!list.contains(matcher.group()))
list.add(matcher.group());
}
List<LabelValueBean> tmpList = new ArrayList<LabelValueBean>();
for(int i = 0; i<list.size(); i++){
String index = i+"";
tmpList.add(new LabelValueBean(index,(String)list.get(i)));
}
return tmpList;
}
这个方法可以帮你

Answer 4

啥意思，举个例子

Answer 5

用Pattern.compile( );方法