sed print only matching group

Question

I want to grab the last two numbers  one int  one float  followed by optional whitespace  and print only them   Example   foo bar  lt foo gt  bla 1 2 3 4   Should print   2 3 4   So far  I have the following   sed -n   s    0-9  0-9      t  0-9       t      replacement p     will give me  foo bar  lt foo gt  bla 1 replacement   However  if I try to replace it with group 1  the whole line is printed   sed -n   s    0-9  0-9      t  0-9       t       1 p     How can I print only the section of the line that matches the regex in the group

User · Answer

grep is the right tool for extracting    using your example and your regex   kent   echo  foo bar  lt foo gt  bla 1 2 3 4  grep -o   0-9  0-9      t  0-9       t     2 3 4

User · Answer

I agree with  kent that this is well suited for grep -o  If you need to extract a group within a pattern  you can do it with a 2nd grep    To extract  1 from  xx  0-9   yy    echo  quot aa678bb xx123yy xx4yy aa42 aa9bb quot    grep -Eo  xx 0-9  yy    grep -Eo   0-9    123 4    To extract  1 from  a  0-9   b    echo  quot aa678bb xx123yy xx4yy aa42 aa9bb quot    grep -Eo  a 0-9  b    grep -Eo   0-9    678 9  I generally cringe when I see 2 calls to grep sed awk piped together  but it s not always wrong  While we should exercise our skills of doing things efficiently   quot A foolish consistency is the hobgoblin of little minds quot   and  quot Real artists ship quot

User · Answer

The cut command is designed for this exact situation  It will  cut  on any delimiter and then you can specify which chunks should be output   For instance  echo  foo bar  lt foo gt  bla 1 2 3 4    cut -d     -f 6-7  Will result in output of  2 3 4  -d sets the delimiter  -f selects the range of  fields  to output  in this case  it s the 6th through 7th chunks of the original string  You can also specify the range as a list  such as 6 7

User · Answer

And for yet another option  I d go with awk    echo  foo bar  lt foo gt  bla 1 2 3 4    awk    print   NF-1    NF       This will split the input  I m using STDIN here  but your input could easily be a file  on spaces  and then print out the last-but-one field  and then the last field  The  NF variables hold the number of fields found after exploding on spaces   The benefit of this is that it doesn t matter if what precedes the last two fields changes  as long as you only ever want the last two it ll continue to work

User · Answer

Match the whole line  so add a    at the beginning of your regex  This causes the entire line to be replaced with the contents of the group   echo  foo bar  lt foo gt  bla 1 2 3 4     sed -n   s      0-9  0-9      t  0-9      t       1 p  2 3 4

[regex] sed: print only matching group

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