英文:
Why isn't my code looking for non-letters? (Regular expressions)
问题
import java.util.Scanner;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class Main {
public static void main(String[] args) {
String naming;
Scanner input = new Scanner(System.in);
System.out.print("What is your name: ");
naming = input.nextLine();
input.close();
//**
Pattern pattern = Pattern.compile("[^a-zA-Z]"); // Updated pattern to match non-letters
Matcher matcher = pattern.matcher(naming); // Use the user's input for matching
boolean check = matcher.find();
if (check) {
System.out.println("Invalid name entered");
//**
} else {
System.out.print("continue");
}
}
}
在 ** 之间的部分是我卡住的部分。这个程序应该要求输入你的名字，如果你输入任何不是字母的字符，它会显示"Invalid name entered"。但它只会显示"continue"。我做错了什么？此外，我想在System.out.println("Invalid name entered");行下使用break，但repl.it告诉我"在循环或开关之外不能使用break"。顺便说一下，我必须在这里使用正则表达式。
我不确定该怎么办。Pattern pattern = Pattern.compile(naming); 应该将名称设置为模式，然后 Matcher matcher = pattern.matcher("[^a-zA-Z]"); 应该查找非字母字符。最后，
boolean check = matcher.find();
if (check) {
System.out.println("Invalid name entered");
} else {
System.out.print("continue");
}
应该在找到非字母字符时打印"Invalid name entered"，但它只是忽略了这一点，而是打印"continue"。
英文:
import java.util.Scanner;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
class Main {
public static void main(String[] args) {
String naming;
Scanner input = new Scanner(System.in);
System.out.print("What is your name: ");
naming = input.nextLine();
input.close();
//**
Pattern pattern = Pattern.compile(naming);
Matcher matcher = pattern.matcher("[^a-zA-Z]");
boolean check = matcher.find();
if (check) {
System.out.println("Invalid name entered");
//**
} else {
System.out.print("continue");
}
}
}
Between the ** is the part I'm stuck on. This program is supposed to ask for your name, and if you enter anything that's not a letter, then it will say "Invalid name entered." But instead it just says "continue". What am I doing wrong? Also, I want to make the program break under the System.out.println("Invalid name entered"); line, but repl.it tells me "break cannot be used outside of a loop or switch". I have to use regular expressions here by the way.
I'm not sure what to do. Pattern pattern = Pattern.compile(naming); is supposed to set the name to a pattern, then Matcher matcher = pattern.matcher("[^a-zA-Z]"); is supposed to look for non-letters. Finally,
boolean check = matcher.find();
if (check) {
System.out.println("Invalid name entered");
} else {
System.out.print("continue");
is supposed to print "Invalid name entered" if non-letters are found, but it just ignores that and prints "continue" instead.
答案1
得分: 0
你的正则表达式和被搜索的字符串的顺序是错误的。而不是
Pattern pattern = Pattern.compile(naming);
Matcher matcher = pattern.matcher("^[a-zA-Z]");
你需要像这样：
Pattern pattern = Pattern.compile("^[a-zA-Z]");
Matcher matcher = pattern.matcher(naming);
英文:
You've got the regular expression and the string that's being searched the wrong way round. Instead of
Pattern pattern = Pattern.compile(naming);
Matcher matcher = pattern.matcher("[^a-zA-Z]");
You need something like
Pattern pattern = Pattern.compile("[^a-zA-Z]");
Matcher matcher = pattern.matcher(naming);
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