从字符串字符和数字中提取字母

Question

I have these Strings :

"Turtle123456_fly.me"
"birdy_12345678_prd.tr"

I want the first words of each, ie:

Turtle
birdy

I tried this:

 Pattern p = Pattern.compile("//d");
 String[] items = p.split(String);

but of course it's wrong. I am not familiar with using Pattern .

Answer 1

String s1 ="Turtle123456_fly.me";
String s2 ="birdy_12345678_prd.tr";

Pattern p = Pattern.compile("^([A-Za-z]+)[^A-Za-z]");
Matcher matcher = p.matcher(s1);

if (matcher.find()) {
    System.out.println(matcher.group(1));
}

Explanation: The first part ^([A-Za-z]+) is a group that captures all the letters anchored to the beginning of the input (using the ^ anchor). The second part [^A-Za-z] captures the first non-letter, and serves as a terminator for the letters sequence. Then all we have left to do is to fetch the group with index 1 (group 1 is what we have in the first parenthesis).

Answer 2

Replace the stuff you don't want with nothing:

String firstWord = str.replaceAll("[^a-zA-Z].*", "");

to leave only the part you want.

The regex [^a-zA-Z] means "not a letter", the everything from (and including) the first non-letter to the end is "removed".

See live demo .

Answer 3

也许你应该试试这个\\d+\\w+.*