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

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 <foo> 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

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

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

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

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

[regex] sed: print only matching group

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