CODEWITHSUNDEEP
bashsed
user1488804
user1488804

Reputation: 860

Unable to capture part of a line using sed -n '/.../p'

I'm using sed to store a string in a variable. I'd like to pick out part of one line in a text file, but am unable to do so.

For example, if a line of text from which I'd like to extract the text is:

{foo bar 1.1.2} foo

And I'd like to extract bar 1.1.2

My command looks like: BarVersion=$(sed -n "/bar [.0-9]/p" file.txt)

However, if I then write:

echo $BarVersion

I find that the sed command has saved the whole line in BarVersion, not the just the part I was looknig for. i.e. I get {foo bar 1.1.2} foo

Upvotes: 0

Views: 208

Answers (3)

jthill
jthill

Reputation: 60443

If you're running GNU/something,

sed -nr '/^\{([^ ]*) ([^}]*)\} \1$/ s//\2/p'

If the matching 'foo's isn't a requirement it's

sed -n /^{[^ ]* \([^}]*\)} .*/ s//\1/p'

Upvotes: 0

Svend Hansen
Svend Hansen

Reputation: 3333

You can do it with grep like this:

BarVersion=$(grep -o "bar [.0-9]*" file.txt)

This seems to work for me, but of course it's not using sed, but in my opinion it's a bit simpler.

-o means only output the part of the line that matched.

Upvotes: 1

user1907906
user1907906

Reputation: 

$ echo '{foo bar 1.1.2} foo' | sed -n 's/^.*\(bar [\.0-9]*\)}.*$/\1/p'
bar 1.1.2
  • s/x/y/ substitutes x with y
  • \(\) forms a group which you can reference via \1
  • [\.0-9]* is any number of dots . and digits
  • ^ is the start of the buffer, $ is the end
  • .* is any number of characters

Upvotes: 2

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