This can be easily done if the first letters are latin characters. But this method does not work with non-latin (for example cyrillic) characters. Just press Ctrl+F, enable “Regular Expression” checkbox, and search for ^(.) replace with (Replace All) \u\1 “. matches newline” checkbox has to be unchecked
Try below code hope its helpful to you. refer TextCapitalization Here Capital first letter TextCapitalization.words Capital each letter TextCapitalization.characters Capital first letter of textfield TextCapitalization.sentences Default lowercase letter of textfield TextCapitalization.none Your Widget: TextField( textCapitalization: TextCapitalization.words, decoration: InputDecoration( border: OutlineInputBorder( borderRadius: BorderRadius.circular(10), ), prefixIcon: Icon( Icons.search, ), hintText: ‘Search’, ), ), Result screen using TextCapitalization.words: … Read more
Here is another go at it: NSString *txt = @”hi my friends!”; txt = [txt stringByReplacingCharactersInRange:NSMakeRange(0,1) withString:[[txt substringToIndex:1] uppercaseString]]; For Swift language: txt.replaceRange(txt.startIndex…txt.startIndex, with: String(txt[txt.startIndex]).capitalizedString)
Use string.isupper() letters = “asdfHRbySFss” uppers = [l for l in letters if l.isupper()] if you want to bring that back into a string you can do: print “”.join(uppers)
There’s no MySQL function to do that, you have to write your own. In the following link there’s an implementation: MySQL Capitalize Function In order to use it, first you need to create the function in the database. You can do this, for example, using MySQL Query Browser (right-click the database name and select Create … Read more
You can use str.title: df.Column1 = df.Column1.str.title() print(df.Column1) 0 The Apple 1 The Pear 2 Green Tea Name: Column1, dtype: object Another very similar method is str.capitalize, but it uppercases only first letters: df.Column1 = df.Column1.str.capitalize() print(df.Column1) 0 The apple 1 The pear 2 Green tea Name: Column1, dtype: object
Pass a function as the repl argument. The MatchObject is passed to this function and .group(1) gives the first parenthesized subgroup: import re s=”start TT end” callback = lambda pat: pat.group(1).lower() re.sub(r'([A-Z]){2}’, callback, s) EDIT And yes, you should use ([A-Z])\1 instead of ([A-Z]){2} in order to not match e.g. AZ. (See @bobince’s answer.) import … Read more
Flutter has a textCapitalization property for textfields. Set this property to TextCapitalization.sentences or any of the available values eg .characters or .words Like so: TextField( keyboardType: TextInputType.text, **textCapitalization: TextCapitalization.sentences,** style: TextStyle( fontSize: 30.0, color: Colors.black, fontWeight: FontWeight.bold ), )
You can use the regex: ^[A-Z0-9]{3}(?:List)?$ Explanation: ^ : Start anchor [A-Z0-9] : Char class to match any one of the uppercase letter or digit {3} : Quantifier for previous sub-regex (?:List) : A literal ‘List’ enclosed in non-capturing paranthesis ? : To make the ‘List’ optional $ : End anchor See it