Bash string matching

开发者_高级运维I need a single line bash command that takes a piped input and returns true or false depending if it match a string or not..

This seems to work for the true or false part:

test `expr match "Release:11.04" "Release:11.04"` = 13 && echo true || echo false

But I cannot get it to work properly with the other command's output..

What I need would be something like this:

test `expr match "`lsb_release -r`" "Release:11.04"` = 13 && echo true || echo false

So I tried with xargs, but I can't get it to work:

lsb_release -r | xargs -I {} test `expr {} "Release: 11.04"` = 13 && echo True || echo False

Also, if any of you happens to know a shorter way to achieve this, it would be greatly welcome.

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You could use the $(cmd) syntax.

test $(expr match "$(lsb_release -r)" "Release:11.04") = 13 && echo true || echo false

Though on my system there is a tab in the output of lsb_release after ':', so you may want to doublecheck your checks.

And actually after fiddling a bit, I think this would be a nicer way with bash (tested on Debian system):

[[ "$(lsb_release -r)" =~ $'Release:\t6.0.2' ]] && echo true || echo false

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Why not this way:

[ `lsb_release -r | awk '{print $2}'` = "11.04" ] && echo true || echo false

BTW, if you need to use nested `` you can escape inner one with \ like that:

test `expr match "\`lsb_release -r\`" "Release:11.04"`

But it looks ugly. Using $() is more readable, but bit less portable.

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This would be another simpler approach -

lsb_release -r | grep -q "11.04" && echo true || echo false

Basically, this greps the output of lsb_release -r for 11.04. -q in grep will return exit status zero if a match is found, which will echo true. false otherwise.