How to delete all white spaces until first character?

Question

Is there a simple way to delete all white spaces on a line until the first character on that line is met?

An example:

           #a list of comments
                # item 1
                # item 2

And I would like it to become:

#a list of comments
# item 1
# item 2

I know how to delete n characters (ex: 8x) and to repeat the command (.), but how could I do without having to input the number of white spaces?

Answer 1

You can either visually select the lines and use

:'<,'>s/^\s*//

Which means 'substitute all of the whitespaces following the first column of the line by nothing'

Or go on the first line, use

• 0 to go on the first column
• d^ to delete until the first character of the line

And then go to the next line and use the dot command

Answer 2

This works for me.

esc : to enter vim command mode

%s/^\s*//g

Meaning:

%s/<REGEX TO REPLACE>/<REPLACEMENT TEXT>/ for string substitution

g for global (all lines)

Regex matching leading whitespace is ^\s*

Voila

Answer 3

In addition to statox's methods, you can:

• Position the cursor at the beginning of the leading whitespace and type dw
• Position the cursor anywhere in the leading whitespace and type diw
• Position the cursor at the first non-space character of the line and type d0
• Visually select all the lines you want to move left, e.g., by typing V on the first line and moving the cursor to the last line, then executing :left

Update

What I actually usually do in such cases is:

• Visually select all the lines as above, type < to move them left by one shiftwidth, then type . until they're shifted all the way to the left margin.

Answer 4

A few variations on a theme:

Method(s) that will work in vi:

• Go to the first line that you want to manipulate.
• Count the lines that you want to manipulate — let’s say there are 17 of them.  Type 17<<.  This will shift each of the following seventeen lines left by one shiftwidth (normally eight characters; i.e., one tab).  Assuming you still have lines with leading spaces, type . to repeat the shift command.  Type . repeatedly until all the spaces are gone.
• If you want to remove the leading spaces from all lines to the end of the buffer, use <G and then . repeatedly.
• Any common technique for identifying the end of the range will work similarly.  For example, with the text in your question, you could use </2/ (and then . repeatedly).

Method that will work only in vim:

• Go to the first line that you want to manipulate.
• Type V or Ctrl+V to go into visual selection mode (mark the beginning of the range).
• Move to the end of the range.
• Type 9<.  This will shift the selected lines left by nine shiftwidths.  As this will typically be 72 characters (9×8), there’s a good chance that that will do the job.  If you still have lines with leading spaces, type . to repeat the shift command (i.e., another72 characters).
• Or, if you know that you have lines with more than 72 leading spaces (or nine tabs), just use 99<.

Answer 5

Deleting a word at the beginning of the line will do the trick:

dw

If you want to repeat that for every line in the file that begins with whitespace:

:g/^ /norm dw

Answer 6

Using vi you should be able to type

<Esc>:%left<CR>

to left-align the whole file.

Answer 7

Here is a single command to do it (starting in Normal mode, on the line you want to modify):

I<c-w><esc>

This switches to Insert mode and positions the cursor on the first non-whitespace character. Ctrl-w will then delete to the beginning of the line.

This can then be . repeated on any other line, from any column.

Answer 8

In normal mode i use d/\w<CR> I often find my self using / or ? to make big surgical jumps/deletions/selections

Answer 9

I've written a function for this (add this in your .vimrc) :

function RTS()
"Remove trailing spaces in every line"
    let nl = line('$')
    exe ":normal gg"
    for i in range(1, nl)
        exe ":normal 0d^j"
    endfor
endfunction

then call it whenever you want to use it:

:call RTS()

Answer 10

Not answering for this specific task, but I personally prefer df<char> and dt<char> or with c operator.

Once you are familiar with it, it deals with most cases more consistently by deleting from your cursor position to a specified character (f for inclusive, t for exclusive) at the cost of only one more keystroke than the optimal d^ and involving no complex mind calculation (IMHO, you'd better avoid F and T for backward search which do not include the character under cursor, this places more burden on memory and mind).

e.g. some of my often used commands: dt) df, cf etc. You only need to count characters or resort to Visual Mode (also remember using f or t to move cursor!) when there are multiple occurrences of the character, which seems to be a rare case for me. Also often we want to delete to a delimiter, such as , ; , which stands out clearly in the text, making obviously the argument of the command and even aiding quickly counting occurrences.