How to wrap this output in quotes?

I have the following command, which gives me 99% of what I want:

root@CA2UA5232QPZ:/# tail -3 newtag | awk '{print $1}'
v1.0.20170512.1
v1.0.20170712.1
v1.0.20170712.2
root@CA2UA5232QPZ:/#

But I need to tweak it so the output looks like this:

'v1.0.20170512.1'
'v1.0.20170712.1'
'v1.0.20170712.2'

0

4 Answers

You could pipe your output to sed:

tail -3 newtag| awk '{print $1}'| sed "s/^/'/;s/$/'/"

2

Another way, using hexadecimal ASCII code for ':

tail -3 newtag | awk '{print "\x27" $1 "\x27" }'

3

There's the hideous shell quoting hell of these

awk '{print "'"'"'" $1 "'"'"'"}'
awk '{print "'\''" $1 "'\''"}'

or use an odd output field separator

awk -v OFS="'" '{print "", $1, ""}'

but this isn't too bad

awk -v q="'" '{print q $1 q}'

1. One pass with sed:

tail -3 newtag | sed "s/\(.*\)\s.*/'\1'/"

Explanation: in sed the replacement string does not have to be quoted

2. Or you could use perl:

tail -3 newtag | perl -anE "say qq('\$F[0]')"

Explanation:

• perl scripting language that excels at text processing
• -a split each line in fields
• n do not automatically print each line
• E execute the following command, and enable features such as say
• " start instructions with " because we later want to use '
• say print the following expression and a newline
• qq( start of literal expression that allows for variable interpolation
• ' a literal '
• \$F[0] the first "field" on the line
• ' a literal '
• ) end of literal expression
• " end of instructions

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