Limit the number of words with regular expression

Question

The regular expression which you gave: ^(?:\\b\\w+\\b[\\s\\r\\n]*){1,250}$ to limit 250 words over multiple lines works if it doesn't have any special characters.

What should I do if I need to search for number of words which also consists special characters? Something like this an example:

--> Hi! i need help with regular expression, please help me. <--

Answer 1

最简单的方法是对单词字符进行分组，并将这些分组限制在特定范围内（1-250）：

^\W*(\w+(\W+|$)){1,250}$

Answer 2

I am not familiar with C# so I will describe the regex.

Method 1:

You are basically looking for this:

(\b[^\s]+\b){1,250}

In java:

\\s is any whitespace character.

[^\\s]+ is a sequence of non-whitespace characters.

\\b is a word boundary.

You can translate the regex to C#.

Method 2:

Tokenize the input text into whitespace delimited words. In java, this is done by:

String[] tokens = inputString.split("\\s+");

where the regex is \\s+

Now you can count the length of the array and implement your logic to reject the words beyond 250.

Method 3:

Define a pattern to capture whitespace as a 'capturing group'.

(\s+)

Now you can do a count the number of matches in your pattern matcher using a while loop. This is essentially kinda same as Method 2 but without involving the creation of the array of tokens.

Answer 3

A bit late to answer but none of the solutions here worked:

^([a-zA-Z0-9]+[^a-zA-Z0-9]*){1,8}$

where {1,8} defines how many wordt you want

Answer 4

You can use the {a,b} quantifiers on any expression, like so:

.{1,256}
[\d\w_?]{1,567}
(0x)?[0-9A-F]{1,}

So, in your case, you could use:

^(?:\b\w+\b[_!?\s\r\n]*){1,250}$

Where the _!? can be any special characters.