Shell scripting. Command substitution issue in my script

There's no obligation to use / as the separator for sed.

s/a/c/

May become

s#a#c#

So in your case:

someidformatted=`echo "${someid}" | sed 's#\/#\\\/#'`

would do the job.

I can only guess that the problem was caused by some lack of / escaping.

Here's what is going on. From the bash(1) man page, emphasis mine:

When the old-style backquote form of substitution is used, backslash retains its literal meaning except when followed by $, ‘, or \. The first backquote not preceded by a backslash terminates the command substitution. When using the $(command) form, all characters between the parentheses make up the command; none are treated specially.

So most likely you need more backslashes for the command substitution than a plain command.You can debug this by setting set -x:

# someidformatted=`echo "${someid}" | sed 's/\//\\\//'`++ echo 314-12345/08++ sed 's/\//\\//'sed: 1: "s/\//\\//": bad flag in substitute command: '/'+ someidformatted=# someidformatted=$(echo "${someid}" | sed 's/\//\\\//')++ echo 314-12345/08++ sed 's/\//\\\//'+ someidformatted='314-12345\/08'

So, you can see that an occurrence of \\ gets turned to \. Adding more backslashes works, but I prefer the $(command) form:

# someidformatted=$(echo "${someid}" | sed 's/\//\\\//')