Unix sed command to find and replace a pattern

Question

I am trying to match the below pattern which has spaces in between.

         -DsomeArg=some value

and to replace it with below pattern using sed command.

         -DsomeArg="some value"

There can be any number of spaces in between. I tried below command but it's not working.

          sed 's/^\(-D.*\)=(.\+\s\+.\+)/\1="\2"/' test.dat

where as .* works instead of .+ , but I want to match one or more pattern. I am not able to find what I am doing wrong.

What is the end of the value? End of line?

theglauber
– theglauber

2013-01-15 19:09:44 +00:00
Commented Jan 15, 2013 at 19:09 — theglauber
– theglauber, Commented Jan 15, 2013 at 19:09
What's wrong with sed 's/\(-D.*=\)\(.*\)/\1"\2"/'?

Beta
– Beta

2013-01-15 19:09:46 +00:00
Commented Jan 15, 2013 at 19:09 — Beta
– Beta, Commented Jan 15, 2013 at 19:09

Adam Sznajder · Accepted Answer · 2013-01-15 19:09:52Z

1

cat /tmp/test | sed -r 's/^(-D.*?)=(.+\s+.+)/\1="\2"/'
-DsomeArg="some value"

Hope that will help :)

answered Jan 15, 2013 at 19:09

Adam Sznajder

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Comments

Rob Davis · Accepted Answer · 2013-01-16 06:29:06Z

0

This does what you asked for:

sed 's/=\(.*\)/="\1"/'

or a little shorter:

sed 's/=/="/;s/$/"/'

But I have a feeling there's more to the story than you're saying. Like, are you expecting some text to follow "some value" that you want to leave outside of the quotation marks? Does something precede -DsomeArg that you don't want to match?

With regular expressions, context is everything.

answered Jan 16, 2013 at 6:29

Rob Davis

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