正则表达式 - 如何识别字符串+空格+字符串

Question

I need to recognize some pattern which goes like this: 我需要认识到一些这样的模式：
[letters][some spaces][letters] [信件] [一些空格] [信件]

What I done so far is this: 到目前为止我做的是：

String regex = "[a-zA-Z]\\s+[a-zA-Z]";

Answer 1

As per the requirement, you wrote letters (with a s at the end). 根据要求，你写了字母（最后用s ）。

[letters][some spaces][letters]

So to do that you must be quantifying the character class as 所以要做到这一点，你必须量化字符类

String regex = "[a-zA-Z]+\\s+[a-zA-Z]+";

[a-zA-Z]+ Matches one or more letters. [a-zA-Z]+匹配一个或多个字母。 Here + is the quantifier which quantifies [a-zA-Z] One or more times. 这里+是量化[a-zA-Z]一次或多次的量词。
Regex Demo 正则表达式演示
Where as if you write [a-zA-Z]\\\\s+[a-zA-Z] , it would only match a single character before and after the space. 就像你写[a-zA-Z]\\\\s+[a-zA-Z] ，它只匹配空格前后的单个字符。
Regex Demo 正则表达式演示

If you want the entire string to follow this pattern, you must be adding anchors as well to the pattern as 如果您希望整个字符串遵循此模式，则必须将锚点添加到模式中

String regex = "^[a-zA-Z]+\\s+[a-zA-Z]+$";

^ Anchors the regex at the start of the string. ^在字符串的开头处锚定正则表达式。
$ Anchors the regex at the end of the string. $在字符串末尾锚定正则表达式。
These anchors ensure that immediatly following start of string, ^ number of letters occure, [a-zA-Z]+ followed by space and again letters. 这些锚点确保立即跟随字符串的开始， ^出现字母数量， [a-zA-Z]+后跟空格和字母。 The second group of letters is followed by end of string $ 第二组字母后跟字符串$结尾