If you want to modify the String, use retain. This is likely the fastest way when available. fn remove_whitespace(s: &mut String) { s.retain(|c| !c.is_whitespace()); } If you cannot modify it because you still need it or only have a &str, then you can use filter and create a new String. This will, of course, have … Read more
You can use a unicode literal for a non breaking space: var foo = document.createTextNode(“\u00A0”);
Use list comprehension — simpler, and just as easy to read as a for loop. my_string = “blah, lots , of , spaces, here ” result = [x.strip() for x in my_string.split(‘,’)] # result is [“blah”, “lots”, “of”, “spaces”, “here”] See: Python docs on List Comprehension A good 2 second explanation of list comprehension.
The solution below works very well for me. Only 4 lines and works with most (all?) characters. Solution: :Trim SetLocal EnableDelayedExpansion set Params=%* for /f “tokens=1*” %%a in (“!Params!”) do EndLocal & set %1=%%b exit /b Test: @echo off call :Test1 & call :Test2 & call :Test3 & exit /b :Trim SetLocal EnableDelayedExpansion set Params=%* … Read more
scala.xml.Utility.trim() should do what you want: scala> val x = <foo> | <bar>hello world</bar> | <baz> xxx </baz> | </foo> x: scala.xml.Elem = <foo> <bar>hello world</bar> <baz> xxx </baz> </foo> scala> scala.xml.Utility.trim(x) res0: scala.xml.Node = <foo><bar>hello world</bar><baz>xxx</baz></foo>
In case you still need to know how to do this. Edit the Registry Key in HKEY_CURRENT_USER: Software\Microsoft\Microsoft SQL Server\xx\Tools\Shell\Text Editor\Visible Whitespace Where xx is the version you have (90 is 2005, 100 is 2008) And set it to 1 to show. Or for SQL Server 2012 and up client tools it is also in … Read more
Like this: str_repeat(‘ ’, 5); // adds 5 spaces