Add an option to string collect to preserve everything except the final newline

[IsaacOscar]

Bassically, consider the following code:

program $variable

This works exactly as I would expect.
But suppose I first want to replace all occurences of foo with bar before calling program, naturally I tried this:

program (string replace -a "foo" "bar" $variable)

Now it works most of the time, but not if $variable contains a newline, in which case it actually passes multiple arguments to foo, none of which contain newlines.

So I tried using string collect:

program (string replace -a "foo" "bar" $variable | string collect)

Which almost works! The newlines are preserved, unless $variable ends in a newline, in which case it is striped.

So next I tried using the --no-trim-newlines option to string collect:

program (string replace -a "foo" "bar" $variable | string collect -N)

Unfortunately now program always gets an extra newline (since string replace itself keeps outputing one). I want the argument to program to end in a newline if and only if $variable does.

Eventuallly, I found a solution that almost works:

program (string replace -a "foo" "bar" $variable | head -c -1 | string collect -N)

Another solution is to use something other than string replace:

program (echo -n $variable | sed 's/foo/bar/g' | string collect -N)

Unfortunatly that doesn't work if $variable is the empty string, because then program get's no arguments. I also have to pass the --allow-empty option to string collect.

My point is that I think it would be very helpfull if string collect had an option that did this (and/or an option to string replace and other builtins to not print newlines I never asked for).

[lavafroth]

You could try something like this:

program "$(string replace "foo" "bar" $variable)"

Here's what I tried:

Setting the variable:

set variable "this is some example text
with some newlines
hehe"

Replacing "some" with "THE":

printf "$(string replace "some" "THE" $variable)"