Regex 只保留字母和空格的正则表达式?
这个正则表达式适用于单词变量,但它删除了定义变量中不需要的所有空白。很难理解为什么Regex 只保留字母和空格的正则表达式?,regex,Regex,这个正则表达式适用于单词变量,但它删除了定义变量中不需要的所有空白。很难理解为什么 String word = "cheese[1]" String definition = "[:the quality of being too obviously sentimental]" String regex = "[^A-Za-z]+"; // how to get it to exclude whitespace? finaldefinition = finaldefinitio
String word = "cheese[1]"
String definition = "[:the quality of being too obviously sentimental]"
String regex = "[^A-Za-z]+"; // how to get it to exclude whitespace?
finaldefinition = finaldefinition.replaceAll(regex,"")
输出:
cheese
the quality of being too obviously sentimental
单词=奶酪
定义=明显具有真实性的质量
预期结果:
单词=奶酪
定义=过于明显的多愁善感
谢谢你抽出时间 你在找这个吗
public static void main(String[] args) {
String[] strs = new String[] {"cheese[1]", "[:the quality of being too obviously sentimental]"};
for (String m: strs){
System.out.println(m.replaceAll("[^a-zA-Z ]", ""));
}
输出:
cheese
the quality of being too obviously sentimental
您可以将
[^A-Za-z]+
用于:
- 任何字符,除了
- 大写字母
- 小写字母
- 空白
[^A-Za-z\s]+
用于:
- 任何字符,除了
- 大写字母
- 小写字母
- 空白(空白、制表符、换行符)
第三种选择:首先将
\s
替换为
,这样制表符和换行符将变为空白或[^A-Za-z\s]+
,如果您想保留制表符和其他形式的空白。哈哈,谢谢。就这么简单吧/掌纹