I need to get only first two numbers of "uname -r" command in bash
example of regular out put:
uname -r
3.5.0-18-generic
what I expect using magic bash options:
3.5
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3 Answers
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assuming you want everything before the second dot, this will do what you want:
uname -r | cut -d. -f1-2
uname itself does not support cutting the output, afaik. The pipe through cut will show you fields 1 and 2 (-f1-2), delimited by dots (-d.)
Yefim Dinitz
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uname -r | sed 's/\([0-9]\+\.[0-9]\+\)\..*/\1/'
dimir
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and do you know how can I capture this into a variable ? – Ba7a7chy Nov 12 '12 at 09:29
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I admit that using Yefim's solution is more efficient, it does not require sed. – dimir Nov 12 '12 at 09:29
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@Ba7a7chy VER=$(uname -r | cut -d. -f1-2) – dimir Nov 12 '12 at 09:30
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You could also accomplish this with parameter expansion:
$ r="$(uname -r)"
$ echo ${r%.*}
3.5
${VAR%pat} non-greedily removes pat from the end of VAR. Note that pat is a glob pattern i.e. dot just means "dot" and star means "any-number-of-chars".
Thor
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