Firestore数据库规则中的ID匹配 [英] Id match in firestore database rules

问题描述

我正在制定我的安全规则，但是当我尝试将文档ID与正则表达式匹配时，它将无法正常工作.我尝试使用 matches 函数，但似乎不接受该方法.

即使我尝试使用Firebase模式YYYY-MM-DD(/^(19 | 20)[0-9] [0-9] [-\\/.](0 [1-9] | 1 [012])[-\\/.](0 [1-9] | [12] [0-9] | 3 [01])$ ​​/)来自

我正在尝试检查此模式的 roomId (/^(\\ d){6，}#[a-zA-Z0-9] {65，} $/)

我尝试删除"" ，但它给了我这个错误:不匹配的输入')'期望{'{'，'/'，PATH_SEGMENT}

(我知道正则表达式可以，但是我不知道为什么它不能在我编写的代码中使用)

解决方案

您正在将Realtime Database和Firestore之间的语法混淆.

在实时数据库安全规则中，正则表达式特定为JavaScript正则表达式，因此将其包含在/中以进行打开和关闭.

在Firestore安全规则中，正则表达式需要以字符串形式传递，这也意味着不应将其包装在/符号中.

所以:

 允许创建:if docId.matches("^(19 | 20)[0-9] [0-9] [-\\/.](0 [1-9] | 1 [012])[-\\/.](0 [1-9] | [12] [0-9] | 3 [01])$"); 

I am working on my security rules, but when I try to match the document id with a regex, it doesn't work. I tried to use the matches function, but it doesn't seem to accept the method.

Even when I tried using the Firebase pattern YYYY-MM-DD (/^(19|20)[0-9][0-9][-\\/. ](0[1-9]|1[012])[-\\/. ](0[1-9]|[12][0-9]|3[01])$/) from here, but it didn't work (I tried with 1950-01-01).

I am trying to check roomId for this pattern (/^(\\d){6,}#[a-zA-Z0-9]{65,}$/)

Edit: I tried removing the " " around the regex but it gives me this error: mismatched input ')' expecting {'{', '/', PATH_SEGMENT}

(I know the regex is OK, but I don't know why it won't work in the code I wrote)

解决方案

You're getting the syntax mixed up between Realtime Database and Firestore.

In Realtime Database security rules, the regular expression is specific as a JavaScript regex, so enclosed in / for opening and closing.

In Firestore security rules the regular expression needs to be passed as a string, which also means it shouldn't be wrapped in / symbols.

So:

allow create: if docId.matches("^(19|20)[0-9][0-9][-\\/. ](0[1-9]|1[012])[-\\/. ](0[1-9]|[12][0-9]|3[01])$");

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