Firebase规则正则表达式生日

问题描述

我正在尝试使用以下正则表达式来验证生日:

I am trying to validate birthdays using the following regex:

^(?:(?:31(\/|-|\.)(?:0?[13578]|1[02]))\1|(?:(?:29|30)(\/|-|\.)(?:0?[1,3-9]|1[0-2])\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:29(\/|-|\.)0?2\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:0?[1-9]|1\d|2[0-8])(\/|-|\.)(?:(?:0?[1-9])|(?:1[0-2]))\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$

此正则表达式在在线正则表达式测试器中进行测试时有效，但是当我尝试在我的Firebase规则中使用此正则表达式时，Firebase似乎不接受它.我还尝试将反斜杠加倍，但仍然没有运气.

This regex works when tested in an online regex tester, but when I try to use this regex inside my firebase rules, Firebase seems to not accept it. I also tried doubling my backslashes and still no luck.

这是我的Firebase规则:

This is my firebase rule:

".validate": "newData.isString() && newData.val().matches(/^(?:(?:31(\/|-|\.)(?:0?[13578]|1[02]))\1|(?:(?:29|30)(\/|-|\.)(?:0?[1,3-9]|1[0-2])\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:29(\/|-|\.)0?2\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:0?[1-9]|1\d|2[0-8])(\/|-|\.)(?:(?:0?[1-9])|(?:1[0-2]))\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$/)"

这是我在Firebase上遇到的错误:非法正则表达式，未转义的^，^只能出现在正则表达式的末尾"

This is the error I get on Firebase: "Illegal regular expression, unescaped ^, ^ can only appear at the end of regular expressions"

如何调整此正则表达式使其在Firebase上运行?

How can I tweak this regex to get it working on Firebase?

推荐答案

您需要在此处做两件事:

You need to do two things here:

• 确保所有反斜杠都加倍
• 将所有非捕获组都变为捕获组，然后重新调整后向引用(请注意，应删除多余的捕获组)(请注意，您不能使用 \ 15 作为后向引用，这似乎只是支持1到9个反向引用)
• 重新配置模式，以使字符串锚的 ^ 开头出现在正则表达式的开头，而 $ 仅出现在正则表达式的结尾(否则，您将得到非法的正则表达式例外).这里的操作很容易，因为您的模式是 ^ a1 $ | ^ a2 $ | ^ a3 $ 类型，并且它等于 ^(?: a1 | a2 | a3)$ .

• make sure all backslashes are doubled
• turn all non-capturing groups into capturing ones and re-adjust the backreferences (note that redundant capturing groups should be eliminated) (note you can't use \15 as backreference, it seems only 1 to 9 backreferences are supported)
• re-vamp the pattern so that the ^ start of string anchor appeared at the beginning and $ only at the end of the regular expression (otherwise, you will get the illegal regex exception). It is easy to do here as your pattern is of the ^a1$|^a2$|^a3$ type, and it is equal to ^(?:a1|a2|a3)$.

图案应该看起来像

newData.val().matches(/^((31([-\\/.])(0?[13578]|1[02])\\3|(29|30)([-\\/.])(0?[13-9]|1[0-2])\\6)(1[6-9]|[2-9]\\d)?\\d{2}|29([-\\/.])0?2\\9((1[6-9]|[2-9]\\d)?(0[48]|[2468][048]|[13579][26])|(16|[2468][048]|[3579][26])00)|(0?[1-9]|1\\d|2[0-8])([-\\/.])(0?[1-9]|1[0-2])[-\\/.](1[6-9]|[2-9]\\d)?\\d{2})$/)

请注意，我也将(\/|-| \ ..)转换为([-\/.])(因为字符类比普通字符类更有效并使用单字符替代形式)，并从 [1,3-9] 中删除逗号-看起来像是一个错字，您想匹配 1 或 3 到 9 ，而 在字符类中按字面意义处理.

Note that I also turned (\/|-|\.) into ([-\/.]) (since a character class is more efficient than plain alternation with single-char alternatives) and remove a comma from [1,3-9] - it looks like a typo, you wanted to match 1 or a digit from 3 to 9, I believe, and a , is treated literally in a character class.

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