eclipse Java 正则表达式（无空格或 0-9）
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Java Regex (No White Space or 0-9)
提问by mnmn
Input, via a question, the report owner's first name as a string.
通过问题输入报告所有者的名字作为字符串。
Need a regular expression to check, conditionally, to make sure the first name doesn't contain any numeric characters, numbers between 0 – 9. If it does you must remove it. The first name can not contain any white space either.
需要一个正则表达式来有条件地检查，以确保名字不包含任何数字字符，0 – 9 之间的数字。如果是，您必须将其删除。名字也不能包含任何空格。
do
{
System.out.println("Please enter your FIRST name:");
firstName = keyboard.next();
firstName= firstName.toUpperCase();
}
while( !firstName.matches("^/s^[a-zA-Z]+$/s"));
System.out.println("Thanks " + firstName);
Output
p
Please enter your FIRST name:
p p
Please enter your FIRST name:
Please enter your FIRST name:
回答by Bohemian
You've got your regex muddled up. Try this:
你的正则表达式搞砸了。尝试这个：
while(!firstName.matches("^[^\d\s]+$"));
The regex "^[^\\d\\s]+$"means "non digits or whitespace, and at least one character"
正则表达式的"^[^\\d\\s]+$"意思是“非数字或空格，以及至少一个字符”
回答by Prince John Wesley
As you have firstnamein uppercase and matchesmethod matches the whole input,
[A-Z]+is sufficient.
正如您firstname在大写和matches方法中匹配整个输入一样，
[A-Z]+就足够了。
while( !firstName.matches("[A-Z]+"));
or
或者
while( !firstName.matches("\p{Lu}+"));
回答by Isuru
Try
尝试
firstName = firstName.replaceAll("[\d\s]", "").toUpperCase();
回答by shinds
You can try this. Created it using Rubular.com. The
你可以试试这个。使用Rubular.com创建它。这
Pattern nameRequirment = Pattern.compile("^((?!.[\d\s]).)*$");
while (!nameRequirement.matcher(myString).matches()){
//... prompt for new name
}
回答by Mike S.
If you want to eliminate digits, just force:
如果你想消除数字，只需强制：
\D*
In your matcher
在你的匹配器中
回答by ioreskovic
Try this one: ^[a-zA-Z,.'-]+$. :D
试试这个：^[a-zA-Z,.'-]+$。:D
Also, if you want to try out your regular expressions, rubular.comis a great place for that :D
另外，如果你想尝试你的正则表达式，rubular.com是一个很好的地方：D
You used Scanner#next, instead of Scanner#nextLine.
您使用了 Scanner#next，而不是 Scanner#nextLine。
From Scanner#next JavaDoc:
来自 Scanner#next JavaDoc：
Finds and returns the next complete token from this scanner. A complete token is preceded and followed by input that matches the delimiter pattern.
从此扫描器中查找并返回下一个完整的令牌。一个完整的标记前后是与分隔符模式匹配的输入。
I believe one such delimiter is \s+:D
我相信这样的分隔符之一是\s+：D
import java.util.Scanner;
public class FirstNameParser {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
String firstName = null;
do {
System.out.print("Please enter your FIRST name: ");
firstName = keyboard.nextLine();
firstName = firstName.toUpperCase();
}
while (!firstName.matches("^[a-zA-Z,.'-]+$"));
System.out.println("Thanks " + firstName);
}
}
