extract last section of data from file using linux command

问题

I have the input file (myfile) as:

>> Vi 'x' found in file /data/152.916612:2,/proforma invoice.doc
>> Vi 'x' found in file /data/152.48152834/Bank T.T Copy 12 d3d.doc
>> Vi 'x' found in file /data/155071755/Bank T.T Copy.doc
>> Vi 'x' found in file /data/1521/Quotation Request.doc
>> Vi 'x' found in file /data/15.462/Quotation Request 2ds.doc
>> Vi 'y' found in file /data/15.22649962_test4/Quotation Request 33  zz (.doc
>> Vi 'x' found in file /data/15.226462_test6/Quotation Request.doc

and I need to extract all data after the words "found in file " to have this output:

/data/152.18224487:2,S/proforma invoice.doc
/data/152.916612:2,/proforma invoice.doc
/data/152.48152834/Bank T.T Copy 12 d3d.doc
/data/155071755/Bank T.T Copy.doc
/data/1521/Quotation Request.doc
/data/15.462/Quotation Request 2ds.doc
/data/15.22649962_test4/Quotation Request 33  zz (.doc
/data/15.226462_test6/Quotation Request.doc   

I'm using this

grep "" myfile  |   awk '{print $7" "$8" "$9" "$10}'

it works but not in all situations , i.e. if there are a lot of spaces, latest words are not returned.

Are there other better ways to have the same output ?

回答1:

1st Solution: This should be easily done by awk.

awk '{sub(/.*found in file/,"")} 1'  Input_file

2nd solution: OR more precisely with making it as a field separator itself :)

awk -F"found in file" '{print $2}'  Input_file

3rd solution: With sed:

sed 's/.*found in file\(.*\)/\1/' Input_file

4th solution: Using match function of awk.

awk 'match($0,/found in file/){print substr($0,RSTART+RLENGTH+1)}' Input_file

5th solution: Using grep's \K option(tested with provided samples).

grep -Po "found in file \K.*" Input_file

来源：https://stackoverflow.com/questions/58602952/extract-last-section-of-data-from-file-using-linux-command