How to replace a character with a newline in Emacs?

Question

I am trying to replace a character - say ; - with a new line using replace-string and/or replace-regexp in Emacs.

I have tried the following commands:

• M-x replace-string RET ; RET \n

  This will replace ; with 2 characters: \n.

• M-x replace-regex RET ; RET \n

  This results in the following error (shown in the minibuffer):

  Invalid use of `\' in replacement text.

What's wrong with using replace-string for this task? Is there any other way to do it?

Thanks.

Answer 1

M-x replace-string RET ; RET C-q C-j.

• C-q for quoted-insert,

• C-j is a newline.

Cheers!

Answer 2

There are four ways I've found to put a newline into the minibuffer.

1. C-o

2. C-q C-j

3. C-q 12 (12 is the octal value of newline)

4. C-x o to the main window, kill a newline with C-k, then C-x o back to the minibuffer, yank it with C-y

Answer 3

Don't forget that you can always cut and paste into the minibuffer.

So you can just copy a newline character (or any string) from your buffer, then yank it when prompted for the replacement text.

Answer 4

More explicitly:

To replace the semi colon character (;) with a newline, follow these exact steps.

1. locate cursor at upper left of buffer containing text you want to change
2. Type m-x replace-string and hit RETURN
3. the mini-buffer will display something like this: Replace string (default ^ -> ):
4. Type in the character you want to replace. In this case, ; and hit RETURN
5. the mini-buffer will display something like this: string ; with:
6. Now execute C-q C-j
7. All instances of semi-colon will be replaced a newline (from the cursor location to the end of the buffer will now appear)

Bit more to it than the original explanation says.

Answer 5

Switch to text-mode

M-x text-mode

Highlight block to indent

Indent

C+M \

Switch back to whatever mode..

Answer 6

inline just: C-M-S-% (if binding keys still default) than replace-string^J