恰好匹配 8 位数字的 Java RegEx
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Java RegEx that matches exactly 8 digits
提问by jarosik
I have a simple RegEx that was supposed to look for 8 digits number:
我有一个简单的 RegEx 应该查找 8 位数字：
String number = scanner.findInLine("\d{8}");
But it turns out, it also matches 9 and more digits number. How to fix this RegEx to match exactly 8 digits?
但事实证明，它也匹配 9 位及更多位数字。如何修复此 RegEx 以精确匹配 8 位数字？
For example: 12345678 should be matched, while 1234567, and 123456789 should not.
例如： 12345678 应该匹配，而 1234567 和 123456789 不应该。
采纳答案by Dalorzo
Try this:
试试这个：
\bis known as word boundary it will say to your regex that numbers end after 8
\b被称为词边界，它会告诉你的正则表达式数字在 8 之后结束
String number = scanner.findInLine("\b\d{8}\b");
回答by A.D
I think this is simple and it works:
我认为这很简单并且有效：
String regEx = "^[0-9]{8}$";
^- starts with[0-9]- use only digits (you can also use\d){8}- use 8 digits$- End here. Don't add anything after 8 digits.
^- 以。。开始[0-9]- 仅使用数字（您也可以使用\d）{8}- 使用 8 位数字$- 到此结束。8 位数字后不要添加任何内容。
回答by Wiktor Stribi?ew
Your regex will match 8 digits anywhere in the string, even if there are other digits after these 8 digits.
您的正则表达式将匹配字符串中任何位置的 8 位数字，即使这 8 位数字之后还有其他数字。
To match 8 consecutive digits, that are not enclosed with digits, you need to use lookarounds:
要匹配未用数字括起来的 8 个连续数字，您需要使用lookarounds：
String reg = "(?<!\d)\d{8}(?!\d)";
See the regex demo
查看正则表达式演示
Explanation:
说明：
(?<!\d)- a negative lookbehind that will fail a match if there is a digit before 8 digits\d{8}8 digits(?!\d)- a negative lookahead that fails a match if there is a digit right after the 8 digits matched with the\d{8}subpattern.
(?<!\d)- 如果 8 位数字之前有一位数字，则负向后视将导致匹配失败\d{8}8位(?!\d)- 如果在与\d{8}子模式匹配的 8 位数字之后有一个数字，则匹配失败的负前瞻。
