How to match version of a command using bash “=~”?

问题

luajit -v
LuaJIT 2.1.0-beta3 -- Copyright (C) 2005-2017 Mike Pall. http://luajit.org/

I want to negate match the version part, and got LUAJIT_VERSION="2.1.0-beta3" at the beginning of bash script. I use:

if ! [[ "$(luajit -v)" =~ LuaJIT\s+"$LUAJIT_VERSION".* ]]; then
#rest of code

But it seems not working whether I put $LUAJIT_VERSION between "" or not:

Any part of the pattern may be quoted to force the quoted portion to be matched as a string ... If the pattern is stored in a shell variable, quoting the variable expansion forces the entire pattern to be matched as a string.

Bash docs

Can you tell me what's the correct way to do this task?

回答1:

\s is not a recognized character class in bash; you need to use [[:blank:]] instead:

if ! [[ "$(luajit -v)" =~ LuaJIT[[:blank:]]+"$LUAJIT_VERSION" ]]; then

(The trailing .* isn't necessary, since regular expressions aren't anchored to the start or end of the string.)

---

However, it's not clear your regular expression needs to be that general. It looks like you can use a single, literal space

if ! [[ "$(luajit -v)" =~ LuaJIT\ "$LUAJIT_VERSION" ]];

or simply use pattern matching:

if [[ "$(luajit -v)" != LuaJIT\ "$LUAJIT_VERSION"* ]];

来源：https://stackoverflow.com/questions/49761054/how-to-match-version-of-a-command-using-bash