Re: How to read an integer from the minibuffer

[Gregory Heytings <gregory@heytings.org>]

>
> Cool (honestly). But do show how to use that to read an integer and only 
> an integer ...
>

You mean, how these two answers can be combined?

(defun restricted-read-from-minibuffer (prompt regexp &optional allowed-chars)
   "Read a string from the minibuffer, with additional constraints.
If the input does not match REGEXP, read a string again.
If ALLOWED-CHARS is a string, the only allowed characters are those in
that string."
   (let ((map nil)
 	(string nil))
     (when (stringp allowed-chars)
       (let ((m (make-keymap)))
 	(define-key m [t]
 		    (lambda ()
 		      (interactive)
 		      (message "Character not allowed")))
 	(define-key m (kbd "RET") #'exit-minibuffer)
 	(define-key m (kbd "<return>") #'exit-minibuffer)
 	(define-key m (kbd "C-j") #'exit-minibuffer)
 	(define-key m (kbd "C-g") #'abort-minibuffers)
 	(dolist (c (split-string allowed-chars "" t))
 	  (define-key m c #'self-insert-command))
 	(setq map m)))
     (while (progn
 	     (setq string (read-from-minibuffer prompt nil map))
 	     (when regexp
 	       (unless (string-match regexp string nil t)
 		 (message "Unexpected input.")
 		 (sit-for 1)
 		 t))))
     string))

With this,

(restricted-read-from-minibuffer "Integer? " "^[0-9][0-9]*$" "0123456789")

will read "an integer and only an integer".  And you can wrap it into a 
string-to-number to get the integer itself.

(BTW, it seems that there's no way in Elisp to "expand" a regexp charset, 
e.g. "[0-9]" into "0123456789".  That would make the ALLOWED-CHARS 
argument easier to type in.)