Hi. I need help to extract a particular string in unix.Here's the text i have. 18713440 0 53 Nov 12 17:43 ./t3ix 2342344 3 446 Nov 14 15:23 ./t4 I need to extract date info, e.g Nov 12 As this is always 3 characters, space then 2 numeric, would using regular expression combined with sed be the answer ? Or are there other ways to extract that string. Thanks for your help

Oops. The spaces in the text are not fixed. It's always different for diffent rows. Thanks

Hi: Are the fields always the 4th and 5th? If so, awk will work: awk ' { print $4 $5 } ' data.file Regards, Nails

Thanks for your help Nails ! This was exactly what I was looking for. Had difficulty with the 'cut' cmd, but I guess this is good enough. Actually I'm trying to execute this piece of code so I can traverse through the output easily. set tester_array=(`find . -ls|grep txt| egrep -v "Nov 14|Nov 13|t3st1|t3st2"| awk '{print $8 " " $9 " " $10 " " $11}'`) However I'm getting this error msg. Too many words from `` I'm just puzzled as it seems to work when I type in find . -ls|grep txt| egrep -v "Nov 14|Nov 13|t3st1|t3st2"| awk '{print $8 " " $9 " " $10 " " $11}' Any idea's of how to include white space in between my field's ? Thanks.

Hi: awk has a printf specification similar to the "C" languages printf. That will preserve your whitespace. I'm not certain what you are doing, but it looks like you only want files that are "txt". Something like this is a little more efficient: # all on one line find . -name "*txt*" -print|xargs egrep -v Nov| awk ' { printf("%s %s\n"), $4 ,$5 } ' Regards, Nails