Getting list of commands using regex

Question

I have list of commands where some are having parameters which I need to skip before executing them.

show abc(h2) xyz
show abc(h2) xyz opq(h1)
show abc(h2) xyz <32>
show abc(a,l) xyz [<32>] opq
show abc

Ultimately, the list has different combinations of ( ), <>, [] with plain text commands. I want to separate out all other commands from plain commands like "show abc".

Processing needed on commands :-

(h1), (h2), (a,l) are to be discarded
<32> - is to be replaced with any ip address
[<32>] - is to be replaced with any integer digit

I tried following but resultant file was empty :-

cat show-cmd.txt | grep "<|(|[" > hard-cmd.txt

How can I get the result file which has no plain commands using regex?

Desired output file :-

show abc xyz
show abc xyz opq
show abc xyz 1.1.1.1
show abc xyz 2 opq

jkshah · Accepted Answer · 2013-10-16 06:59:20Z

1

Try using grep followed by sed

grep '[(<\[]' file | sed -e 's/\[<32>\]/2/g' -e 's/<32>/1.1.1.1/g' -e 's/([^)]*)//g'

Output:

show abc xyz
show abc xyz opq
show abc xyz 1.1.1.1
show abc xyz 2 opq

Please note that order of s///g command matters in your case.

Also try avoiding redundant use of cat

answered Oct 16, 2013 at 6:59

jkshah

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nutsiepully · Accepted Answer · 2013-10-16 06:31:40Z

0

cat show-cmd.txt | grep "[\[\(\<]" > hard-cmd.txt

This should work. The opening and closing square brackets [] mean that only one of the options need to be present. Then the further brackets that you want to search for are provided and escaped by a .

Hope this helps. Pulkit

answered Oct 16, 2013 at 6:31

nutsiepully

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