How to delete all lines matching a pattern and a line after in Vim?
You can use :g/word/normal 2dd This finds all instances of word and then executes the command after it. In this case it executes 2dd in normal mode
regex
Extract digits from string – Google spreadsheet
You may replace all non-digit characters using the \D+ regex and an empty string replacement with =REGEXREPLACE(A11,”\D+”, “”) or with casting it to a number: =VALUE(REGEXREPLACE(A11,”\D+”, “”))
Regex match digits, comma and semicolon?
You almost have it, you just left out 0 and forgot the quantifier. word.matches(“^[0-9,;]+$”)
Best equivalent for IsInteger in SQL Server
Late entry that handles negative ISNUMERIC(zipcode + ‘.0e0’) –integer ISNUMERIC(zipcode + ‘e0’) –decimal For more see this
How do I check for vowels in JavaScript?
benchmark I think you can safely say a for loop is faster. I do admit that a regexp looks cleaner in terms of code. If it’s a real bottleneck then use a for loop, otherwise stick with the regular expression for reasons of “elegance” If you want to go for simplicity then just use function … Read more
Add http(s) to URL if it’s not there?
Use a before filter to add it if it is not there: before_validation :smart_add_url_protocol protected def smart_add_url_protocol unless url[/\Ahttp:\/\//] || url[/\Ahttps:\/\//] self.url = “http://#{url}” end end Leave the validation you have in, that way if they make a typo they can correct the protocol.
Regular expression to extract URL from an HTML link
If you’re only looking for one: import re match = re.search(r’href=[\'”]?([^\'” >]+)’, s) if match: print(match.group(1)) If you have a long string, and want every instance of the pattern in it: import re urls = re.findall(r’href=[\'”]?([^\'” >]+)’, s) print(‘, ‘.join(urls)) Where s is the string that you’re looking for matches in. Quick explanation of the … Read more
Using regular expressions to validate a numeric range
Using regular expressions to validate a numeric range To be clear: When a simple if statement will suffice if(num < -2055 || num > 2055) { throw new IllegalArgumentException(“num (” + num + “) must be between -2055 and 2055”); } using regular expressions for validating numeric ranges is not recommended. In addition, since regular … Read more