CODEWITHSUNDEEP
bashsed
user2150250
user2150250

Reputation: 5167

Ignoring lines with blank or space after character using sed

I am trying to use sed to extract some assignments being made in a text file. My text file looks like ...

color1=blue
color2=orange

name1.first=Ahmed
name2.first=Sam
name3.first=
name4.first=
name5.first=
name6.first=

Currently, I am using sed to print all the strings after the name#.first's ...

sed 's/name.*.first=//' file

But of course, this also prints all of the lines with no assignment ...

Ahmed
Sam




# I'm just putting this comment here to illustrate the extra carriage returns above; please ignore it

Is there any way I can get sed to ignore the lines with blank or whitespace only assignments and store this to an array? The number of assigned name#.first's is not known, nor are the number of assignments of each type in general.

Upvotes: 1

Views: 571

Answers (2)

Gilles Quénot
Gilles Quénot

Reputation: 185254

sed -n 's/^name[0-9]\.\w\+=\(\w\+\)/\1/p' file

Output

Ahmed
Sam

Explainations

  • the -n switch suppress the default behavior of sed : printing all lines
  • s/// is the skeleton for a substitution
  • ^ match the beginning of a line
  • name literal string
  • [0-9] a digit alone
  • \.\w\+ a literal dot (without backslash means any character) followed by a word character [a-zA-Z0-9_] al least one : \+
  • ( ) is a capturing group and \1 is the captured group

Upvotes: 1

glenn jackman
glenn jackman

Reputation: 246877

This is a slight variation on sputnick's answer:

sed -n '/^name[0-9]\.first=\(.\+\)/ s//\1/p'

The first part (/^name[0-9]\.first=\(.\+\)/) selects the lines you want to pass to the s/// command. The empty pattern in the s command re-uses the previous regular expression and the replacement portion (\1) replaces the entire match with the contents of the first parenthesized part of the regex. Use the -n and p flags to control which lines are printed.

Upvotes: 2

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