I've written a regular expression that matches any number of letters with any number of single spaces between the letters. I would like that regular expression to also enforce a min and max number of characters, but I'm not sure how to do that (or if it's possible).
我写了一个正则表达式，它匹配任意数量的字母以及任意数量的单个空格。我希望正则表达式也强制执行最小和最大字符数，但我不确定如何执行(如果可能的话)。
My regular expression is:
我的正则表达式是:
[A-Za-z](\s?[A-Za-z])+
I realized it was only matching two sets of letters surrounding a single space, so I modified it slightly to fix that. The original question is still the same though.
我意识到它只匹配了一个空格周围的两组字母，所以我稍微修改了一下。最初的问题仍然是一样的。
Is there a way to enforce a minimum of three characters and a maximum of 30?
是否有一种方法可以强制执行最少三个字符和最多30个字符?
4 个解决方案
#1
30
Yes
是的
Just like + means one or more you can use {3,30} to match between 3 and 30
正如+表示可以使用{3,30}匹配3到30之间的一个或多个
For example [a-z]{3,30} matches between 3 and 30 lowercase alphabet letters
例如[a-z]{3,30}匹配3到30个小写字母
From the documentation of the Pattern class
来自Pattern类的文档
X{n,m} X, at least n but not more than m times
In your case, matching 3-30 letters followed by spaces could be accomplished with:
在你的例子中，3-30个字母和空格的匹配可以通过以下方式完成:
([a-zA-Z]\s){3,30}
If you require trailing whitespace, if you don't you can use: (2-29 times letter+space, then letter)
如果需要拖尾空格，可以使用:(2-29乘以字母+空格，然后是字母)
([a-zA-Z]\s){2,29}[a-zA-Z]
If you'd like whitespaces to count as characters you need to divide that number by 2 to get
如果你想用空格作为字符，你需要将这个数字除以2。
([a-zA-Z]\s){1,14}[a-zA-Z]
You can add \s? to that last one if the trailing whitespace is optional. These were all tested on RegexPlanet
您可以添加\ s ?如果末尾的空格是可选的，则对应最后一个空格。这些都是在RegexPlanet上测试的
If you'd like the entire string altogether to be between 3 and 30 characters you can use lookaheads adding (?=^.{3,30}$) at the beginning of the RegExp and removing the other size limitations
如果你想整个字符串完全是3至30个字符可以使用超前添加(? = ^。{ 3 30 } $)初的RegExp和删除其他大小限制
All that said, in all honestly I'd probably just test the String's .length property. It's more readable.
老实说，我可能只是测试了字符串的。length属性。它是更具可读性。
#2
5
This is what you are looking for
这就是你要找的
^[a-zA-Z](\s?[a-zA-Z]){2,29}$
^ is the start of string
^是字符串的开始
$ is the end of string
$是字符串的末尾
(\s?[a-zA-Z]){2,29} would match (\s?[a-zA-Z]) 2 to 29 times..
(\ s ?[a-zA-Z]){ 2,29 }将匹配(\ s ?[a-zA-Z])2 - 29倍. .
#3
3
Actually Benjamin's answer will lead to the complete solution to the OP's question. Using lookaheads it is possible to restrict the total number of characters AND restrict the match to a set combination of letters and (optional) single spaces.
实际上本杰明的回答将会引出OP问题的完整答案。使用lookaheads可以限制字符的总数，并将匹配限制为一组字母和(可选的)单个空格。
The regex that solves the entire problem would become
解决整个问题的regex将变成
(?=^.{3,30}$)^([A-Za-z][\s]?)+$
This will match AAA, A A and also fail to match AA A since there are two consecutive spaces. I tested this at http://regexpal.com/ and it does the trick.
这将会匹配AAA, A，也不能匹配AAA，因为有两个连续的空格。我在http://regexpal.com/中测试了这个功能，它可以做到这一点。
#4
0
you should use
你应该使用
[a-zA-Z ]{20}
[for allowed characters]{for limit of number of characters}
[for allowed characters]{for limit of characters}
#1
30
Yes
是的
Just like + means one or more you can use {3,30} to match between 3 and 30
正如+表示可以使用{3,30}匹配3到30之间的一个或多个
For example [a-z]{3,30} matches between 3 and 30 lowercase alphabet letters
例如[a-z]{3,30}匹配3到30个小写字母
From the documentation of the Pattern class
来自Pattern类的文档
X{n,m} X, at least n but not more than m times
In your case, matching 3-30 letters followed by spaces could be accomplished with:
在你的例子中，3-30个字母和空格的匹配可以通过以下方式完成:
([a-zA-Z]\s){3,30}
If you require trailing whitespace, if you don't you can use: (2-29 times letter+space, then letter)
如果需要拖尾空格，可以使用:(2-29乘以字母+空格，然后是字母)
([a-zA-Z]\s){2,29}[a-zA-Z]
If you'd like whitespaces to count as characters you need to divide that number by 2 to get
如果你想用空格作为字符，你需要将这个数字除以2。
([a-zA-Z]\s){1,14}[a-zA-Z]
You can add \s? to that last one if the trailing whitespace is optional. These were all tested on RegexPlanet
您可以添加\ s ?如果末尾的空格是可选的，则对应最后一个空格。这些都是在RegexPlanet上测试的
If you'd like the entire string altogether to be between 3 and 30 characters you can use lookaheads adding (?=^.{3,30}$) at the beginning of the RegExp and removing the other size limitations
如果你想整个字符串完全是3至30个字符可以使用超前添加(? = ^。{ 3 30 } $)初的RegExp和删除其他大小限制
All that said, in all honestly I'd probably just test the String's .length property. It's more readable.
老实说，我可能只是测试了字符串的。length属性。它是更具可读性。
#2
5
This is what you are looking for
这就是你要找的
^[a-zA-Z](\s?[a-zA-Z]){2,29}$
^ is the start of string
^是字符串的开始
$ is the end of string
$是字符串的末尾
(\s?[a-zA-Z]){2,29} would match (\s?[a-zA-Z]) 2 to 29 times..
(\ s ?[a-zA-Z]){ 2,29 }将匹配(\ s ?[a-zA-Z])2 - 29倍. .
#3
3
Actually Benjamin's answer will lead to the complete solution to the OP's question. Using lookaheads it is possible to restrict the total number of characters AND restrict the match to a set combination of letters and (optional) single spaces.
实际上本杰明的回答将会引出OP问题的完整答案。使用lookaheads可以限制字符的总数，并将匹配限制为一组字母和(可选的)单个空格。
The regex that solves the entire problem would become
解决整个问题的regex将变成
(?=^.{3,30}$)^([A-Za-z][\s]?)+$
This will match AAA, A A and also fail to match AA A since there are two consecutive spaces. I tested this at http://regexpal.com/ and it does the trick.
这将会匹配AAA, A，也不能匹配AAA，因为有两个连续的空格。我在http://regexpal.com/中测试了这个功能，它可以做到这一点。
#4
0
you should use
你应该使用
[a-zA-Z ]{20}
[for allowed characters]{for limit of number of characters}
[for allowed characters]{for limit of characters}