Re: I can't believe: replace string in a string (without tampering a buffer) - does not exist

Swami Tota Ram Shankar <[email protected]> writes:

> Subject -  I can't believe: replace string in a string (without
> tampering a buffer) - does not exist
>
> Hello colleagues,
>
> I am simply trying to replace a string by a string in another string.
> This is so that can cook it, ie do replacements in a context without
> having to match/coordinates/narrow/widen steps, and thus check prior
> to running the final replacement. I prefer to be helped with a simple
> lisp function, perhaps a lambda or some nested operations than a
> defun.
>
> I have done my homework before asking for help, and used the function
>         replace-regexp-in-string
> but the regexp is written as a string and can have wild-cards
> characters so I dont like it as compared to a literal string. If there
> is an option in it to make it literal easily without extensive
> quoting, then I am not familiar with it.

Why do you want to avoid extensive quoting?

(defun replace-in-string (new old string)
  (let ((case-fold-search nil))
   (replace-regexp-in-string (regexp-quote old) new string t)))

(replace-in-string "Pascal" "{NAME}" "Hello {NAME}!")
--> "Hello Pascal!"


> replace is an autoloaded Lisp function in `cl-seq'.
> [Arg list not available until function definition is loaded.]
> not documented

It's documented by the Common Lisp standard.
http://www.lispworks.com/documentation/HyperSpec/Front/index.htm
http://www.lispworks.com/documentation/HyperSpec/Body/f_replac.htm#replace


-- 
__Pascal Bourguignon__                     http://www.informatimago.com/
A bad day in () is better than a good day in {}.