java 如何从正则表达式中的模式匹配中排除某些字符？
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How do I exclude certain characters from pattern matching in regex?
提问by lenignes
I'm trying to implement my own version of atoi and so I want to check if my string contains non-numeric characters and handle the error, but I also also want the search pattern to exclude the + and - symbols (i.e + and - at the start of the string only are valid symbols). I currently have word.matches("^[+=][a-zA-Z]+"), but am not sure how to change it accordingly to fit my needs.
Ex: 20e48 is invalid, 204-8 is invalid, +2048 is valid and so is -2048
我正在尝试实现我自己的 atoi 版本，因此我想检查我的字符串是否包含非数字字符并处理错误，但我还希望搜索模式排除 + 和 - 符号（即 + 和 -只有在字符串的开头是有效符号）。我目前有word.matches("^[+=][a-zA-Z]+")，但不确定如何相应地更改它以满足我的需要。例如：20e48 无效，204-8 无效，+2048 有效，-2048 也是如此
采纳答案by Darshan Mehta
Here you go:
干得好：
public static void main(String[] args) {
String pattern = "^[+-]?[0-9]+$";
System.out.println("20e48".matches(pattern));
System.out.println("204-8".matches(pattern));
System.out.println("+2048".matches(pattern));
System.out.println("-2048".matches(pattern));
System.out.println("2048".matches(pattern));
}
It prints:
它打印：
false
false
true
true
true
Explanation:
解释：
^ => Starts
[+-] => Either plus or minus sign
? => Zero or one occurance
[0-9] => Any number
+ => One or more occurance
$ => End
If any string does not match this pattern, it is not a valid input.
如果任何字符串与此模式不匹配，则它不是有效输入。
回答by rentedrainbow
Try ^ followed by character you don't want to match in a square bracket. E.g. [^k] will not match k character in given string.
尝试 ^ 后跟您不想在方括号中匹配的字符。例如 [^k] 将不匹配给定字符串中的 k 个字符。
回答by StvnBrkdll
This regex may work for you:
这个正则表达式可能对你有用：
^[+-]{0,1}[0-9]*[^0-9]+[0-9]*$
It matches any string that (optionally) begins with + or -, followed by 0 or more numeric characters, followed by 1 or more non-numeric characters, followed by 0 or more numeric characters.
它匹配（可选）以 + 或 - 开头、后跟 0 个或多个数字字符、1 个或多个非数字字符、0 个或多个数字字符的任何字符串。
