This should do what you’re looking for: function clean($string) { $string = str_replace(‘ ‘, ‘-‘, $string); // Replaces all spaces with hyphens. return preg_replace(‘/[^A-Za-z0-9\-]/’, ”, $string); // Removes special chars. } Usage: echo clean(‘a|”bc!@£de^&$f g’); Will output: abcdef-g Edit: Hey, just a quick question, how can I prevent multiple hyphens from being next to each … Read more
preg_quote() is what you are looking for: Description string preg_quote ( string $str [, string $delimiter = NULL ] ) preg_quote() takes str and puts a backslash in front of every character that is part of the regular expression syntax. This is useful if you have a run-time string that you need to match in … Read more
Oh Yes You Can Use Regexes to Parse HTML! For the task you are attempting, regexes are perfectly fine! It is true that most people underestimate the difficulty of parsing HTML with regular expressions and therefore do so poorly. But this is not some fundamental flaw related to computational theory. That silliness is parroted a … Read more
Quick answer: ^[\w*]$ will match a string consisting of a single character, where that character is alphanumeric (letters, numbers) an underscore (_) or an asterisk (*). Details: The “\w” means “any word character” which usually means alphanumeric (letters, numbers, regardless of case) plus underscore (_) The “^” “anchors” to the beginning of a string, and … Read more
^(\d{3}|\d{6})$ You have to have some sort of terminator otherwise \d{3} will match 1234. That’s why I put ^ and $ above. One alternative is to use lookarounds: (?<!\d)(\d{3}|\d{6})(?!\d) to make sure it’s not preceded by or followed by a digit (in this case). More in Lookahead and Lookbehind Zero-Width Assertions.
You can construct a new character class: [^\W\d_] instead of \w. Translated into English, it means “Any character that is not a non-alphanumeric character ([^\W] is the same as \w), but that is also not a digit and not an underscore”. Therefore, it will only allow Unicode letters.
Regex lineSplitter = new Regex(@”[\s*\*]*\|[\s*\*]*”); var columns = lineSplitter.Split(data).Where(s => s != String.Empty); or you could simply do: string[] columns = data.Split(new char[] {‘|’}, StringSplitOptions.RemoveEmptyEntries); foreach (string c in columns) this.textBox1.Text += “[” + c.Trim(‘ ‘, ‘*’) + “] ” + “\r\n”; And no, there is no option to remove empty entries for RegEx.Split as … Read more
The first one won’t store the capturing group, e.g. $1 will be empty. The ?: prefix makes it a non capturing group. This is usually done for better performance and un-cluttering of back references. In the second example, the characters in the capturing group will be stored in the backreference $1. Further Reading.
Use a look-behind: (?<=:)[\w+.-]+ A look-behind (coded as (?<=someregex)) is a zero-width match, so it asserts, but does not capture, the match. Also, your regex may be able to be simplified to this: (?<=:)[^\]]+ which simply grabs anything between (but not including) a : and a ]
For some reason, the above answer didn’t work for me. For those like me who come after, here is what I found. I was expecting a single backslash to escape the bracket, however, you must use two if you have the pattern stored in a string. The first backslash escapes the second one into the … Read more