In BASH, how does ||: get interpreted. I know || is logical or. And I believe : evaluates to true. Can someone give a thorough explanation for this usage?

Example

for i in $IGGY
	do
	    [ "$db" == "$i" ] && skipdb=1 || :

&& and || are short-form if-then's.

true && echo "this will print"
false && echo "But this won't"

true || echo "This won't print"
false || echo "But this will"

$ true && echo success || echo failure

succcess

$ false && echo success || echo failure

failure

$

Thanks for your answer. I have seen this in more than one bash script, so I assume it perhaps had a purpose in a prior version of the shell. Perhaps?

You see many strange and redundant things in copied shell scripts, many of which do worse than nothing. People fiddle until it 'works' without really knowing the language.

This does literally nothing, but at least seems harmless.

Given the 3 line shell script fragment:

for i in $IGGY
	do
	    [ "$db" == "$i" ] && skipdb=1 || :

: evaluates in POSIX shells to the result of /usr/bin/true or true.

It is used like that to override a return status value that would affect subsequent processing.

For example, in Appworx (a batch control system) a false return code almost anywhere is detectable, so that if ||: were not there at the end of a line, appworx may see a false return and abort the job when the child process returns. Even though the job may have completed correctly.

Don't ask how, I'd have to guess.

A likely way to make this happen would be to use the command:
set -e
in a script that Appworx uses to start jobs, or for Appworx to use sh -e when it invokes a shell script.