I am trying to use awk command to find line(s) which the third columns is not digit/date? Suppose there is a file comma , field separated, has three columns and as code , measure , dd/mm/yyyy,97xx574,26.7,12/30/1997,97xy575,18,12/30/1997,code,meas,EXAMDATE,B529ui,28.2,12/30/1997,B530sx,26.4,12/30/1997,J487sxv,21.5,12/30/1997,I tried:awk -F , '/$3[^0-9].*/ {print $0}' <filename> ... but apparently it is not correct!Any help with explanations is highly appreciated it as I am new to Linux. 
How I can find line(s) which the third columns is not digit/date?
linux;shell script;command line;awk
How about this.  Where 3rd field does not consist of 0-9 or /, print the line (which is the default action : no need for a print $0.$3 = third field!~ = where does not (!) match regular expression/  = mark start of regular expression^  = match start of field[0-9/]+ = match any of the 0123456789/ characters at least once$  = match end of field/  = mark end of regular expressionSo code, with output:awk -F, '$3!~/^[0-9/]+$/' filenamecode,meas,EXAMDATE,Introducing a bit more checking, so has to consist of nn/nn/nnnn, try this.awk -F, '$3!~/^[0-9][0-9]\/[0-9][0-9]\/[0-9][0-9][0-9][0-9]$/' filenamecode,meas,EXAMDATE,Could even use grep if you wanted.grep -vE '^.*,.*,[0-9/]+,$' x1code,meas,EXAMDATE,