use*_*595 22 java string removeall
我已经在这个社区的帮助下取代了从用户输入中删除数值,但是,我的代码只会在删除数字之前检索字母字符:
import java.util.Scanner;
public class Assignment2_A {
public static void main(String[] args) {
Scanner firstname = new Scanner(System.in);
String firstname1 = firstname.next();
firstname1 = firstname1.replaceAll("[^A-Z]","");
System.out.println(firstname1);
}
}
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例如,如果用户输入= S1234am,我只会回来:S.如何检索字符串中的剩余字符?
jlo*_*rdo 44
这将删除所有数字:
firstname1 = firstname1.replaceAll("\\d","");
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Vis*_*l K 21
您可以使用:
firstname1 = firstname1.replaceAll("[0-9]","");
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这将从String中删除所有数值firstName1
.
String firstname1 = "S1234am";
firstname1 = firstname1.replaceAll("[0-9]","");
System.out.println(firstname1);//Prints Sam
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如何从字符串中删除数值:
这样做就足够了
str.replaceAll("[^A-Za-z]","");
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但是如果您的字符串包含以下字符怎么办:
String str = "stackoverflow elenasys +34668555555 # ?????????? ?????????? ????? ????? ???? ???? ki?m tra ki?mtra ??????? ??????? ?????? ?? ?????? ??";
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大多数字符也将被删除,所以这是一个更好的选择:
str = str.replaceAll("[\\d.]", "");
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删除所有数值并获得结果:
stackoverflow elenasys + # ?????????? ?????????? ????? ????? ???? ???? ki?m tra ki?mtra ??????? ??????? ?????? ?? ?????? ??
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