正则表达式 (java) 匹配等于或大于 20、递增 5、不允许前导零的数字 [英] regex (java) match numbers equal or greater than 20, incremented by 5, no leading zeros allowed

问题描述

我是新用户，如果这个问题应该在原帖中回答，请原谅:正则表达式匹配大于或等于 20 的数字，以 5 为增量，范围为 20 到 999

I am a new user, so excuse if this question should be answered in it's original post: regex match numbers greater or equal than 20, by increments of 5, range 20 to 999

然而，这是当前的问题:

Neverteless, here is the current question:

• 匹配所有等于或大于 20 的数字(无上限)
• 以 5 为增量
• 无小数点
• 不应允许前导零

使用 stackoverflow 用户 YMI 对另一篇文章的回应:

With stackoverflow user YMI response on another post:

(\d{2}|[2-9])[05]

和用户 nhahtdh

^([2-9]|[1-9]\d)[05]$

但是我想探索没有上限的选项，也不允许前导零.

However I would like to explorer the option of not having upper limit and the leading zeros not being allowed also.

推荐答案

我的回答和nhahtdh's 但请注意 \d+ 而不是 \d ，它对字符数没有上限.

My answer is very similar to nhahtdh's but note the \d+ rather than a \d which places no upper limit on the number of characters.

你会想要一个这样的正则表达式:

You'll want a regex like this:

\b((?:[23456789]|[123456789]\d+)[05])\b

[现场示例]

快速解释这里发生的事情:

To give a quick explanation of what's happening here:

• \b 匹配边界，如空格或符号，以便 \bs 会从文本中找到完整的单词
• 接下来我们为单词前缀提供两个选项，它可以是一个大于等于 2 的数字:[23456789]
• 或者可以是 2 个或多个不以 0 开头的数字:[123456789]\d+
• 对于我们的后缀，我们要求它是 5 的倍数:[05]

• \b matches a boundary, like a white-space or a symbol so the \bs will find complete words from the text
• Next we give two options for the word prefix, it can be a single number 2 or greater: [23456789]
• Or it can be 2 or more numbers that are not led by a 0: [123456789]\d+
• For our suffix we require it to be a multiple of 5: [05]

顺便说一句，您可以通过简单地消耗 0 然后匹配数字来匹配满足其他条件但也具有前导 0 的数字，注意添加 0* 它将匹配任意数量的前导 0:

Incidentally you can match numbers that meet your other criteria but also posses leading 0s by simply consuming the 0s and then matching the number, note the addition of 0* which will match any number of leading 0s:

\b0*((?:[23456789]|\d{2,})[05])\b

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