How can I match the number of words in a string to be > then 5 using regex?
如何使用正则表达式匹配字符串中的单词数> 5然后?
Input1: stack over flow => the regex will not match anything
Input1:堆栈溢出=>正则表达式不匹配任何东西
Input2: stack over flow stack over => the regex will match this string
Input2:堆栈超过流堆栈=>正则表达式将匹配此字符串
I have tried counting the spaces with /\/s/ but that didn't really helped me, because I need to match only strings with no of words > 5
我已经尝试使用/ \ / s /计算空格,但这并没有真正帮助我,因为我只需要匹配没有单词> 5的字符串
Also I don't want to use split by spaces.
另外我不想使用空格分割。
2 个解决方案
#1
4
I would rely on a whitespace/non-whitespace patterns and allow trailing/leading whitespace:
我将依赖于空白/非空白模式并允许尾随/前导空格:
^\s*\S+(?:\s+\S+){4,}\s*$
See demo
Explanation:
-
^- start of string -
\s*- optional any number of whitespace symbols -
\S+- one or more non-whitespace symbols -
(?:\s+\S+){4,}- 4 or more sequences of one or more whitespace symbols followed with one or more non-whitespace symbols -
\s*- zero or more (optional) trailing whitespace symbols -
$- end of string
^ - 字符串的开头
\ s * - 可选任意数量的空白符号
\ S + - 一个或多个非空白符号
(?:\ s + \ S +){4,} - 一个或多个空白符号的4个或更多序列,后跟一个或多个非空白符号
\ s * - 零个或多个(可选)尾随空格符号
$ - 结束字符串
#2
5
^ *\w+(?: +\w+){4,}$
You can use this regex.See demo.
你可以使用这个regex.See演示。
#1
4
I would rely on a whitespace/non-whitespace patterns and allow trailing/leading whitespace:
我将依赖于空白/非空白模式并允许尾随/前导空格:
^\s*\S+(?:\s+\S+){4,}\s*$
See demo
Explanation:
-
^- start of string -
\s*- optional any number of whitespace symbols -
\S+- one or more non-whitespace symbols -
(?:\s+\S+){4,}- 4 or more sequences of one or more whitespace symbols followed with one or more non-whitespace symbols -
\s*- zero or more (optional) trailing whitespace symbols -
$- end of string
^ - 字符串的开头
\ s * - 可选任意数量的空白符号
\ S + - 一个或多个非空白符号
(?:\ s + \ S +){4,} - 一个或多个空白符号的4个或更多序列,后跟一个或多个非空白符号
\ s * - 零个或多个(可选)尾随空格符号
$ - 结束字符串
#2
5
^ *\w+(?: +\w+){4,}$
You can use this regex.See demo.
你可以使用这个regex.See演示。