使用regex检查字符串中单个字符的特定数目。

I'm trying to create a regex pattern for my powershell code. I've never worked with regex before, so I'm a total noob.

我正在为我的powershell代码创建一个regex模式。我以前从来没有和regex合作过，所以我是一个新手。

The regex should check if there are two points in the string.

regex应该检查字符串中是否有两个点。

Examples that SHOULD work:

例子应该工作:

3.1.1
5.10.12
10.1.15

Examples that SHOULD NOT work:

不应该起作用的例子:

3
3.1
5.10.12.1

The string must have two points in it, the number of digits doesn't matter.

字符串中必须有两个点，数字的数量无关紧要。

I've tried something like this, but it doesn't really work and I think its far from the right solution...

我尝试过这样的方法，但并不奏效，我认为这远远不是正确的解决方法。

([\d]*.[\d]*.[\d])

2 个解决方案

#1

In your current regex I think you could escape the dot \. or else the dot would match any character.

在您当前的regex中，我认为您可以避免使用点\。否则点会匹配任何字符。

You could add anchors for the start ^ and the end $ of the string and update your regex to ^\d*\.\d*\.\d*$ That would also match ..4 and ..

您可以添加锚开始^和$的字符串和更新你的regex ^ \ d * \ \ d * \。还可以匹配……4 . .

Or if you want to match one or more digits, I think you could use ^\d+(?:\.\d+){2}$

或者如果你想匹配一个或多个数字,我认为你可以用^ \ d +(?:\ \ d +){ 2 } $

That would match

这将匹配

^       # From the beginning of the string
\d+     # Match one or more digits
(?:     # Non capturing group
  \.\d+ # Match a dot and one or more ditits
){2}    # Close non capturing group and repeat 2 times
$       # The end of the string

#2

Use a lookahead:

使用一个超前:

^\d(?=(?:[^.]*\.[^.]*){2}$)[\d.]*$

Broken down, this says:

^                       # start of the line
\d                      # at least one digit
(?=                     # start of lookahead
    (?:[^.]*\.[^.]*){2} # not a dot, a dot, not a dot - twice
$                       # anchor it to the end of the string
)
[\d.]*                  # only digits and dots, 0+ times
$                       # the end of the string

See a demo on regex101.com.

#1