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I have the date in a CSV file as 19/10/2014. I want it in the format 19-Oct-2014. I am new to Unix so I don't know any command.

A solution completely in GNU awk is as follows:

awk ' split($0,slt,"/");datspec=mktime(slt[3]" "slt[2]" "slt[1]" 00 00 00");print strftime("%d-%b-%Y",datspec) ' <<< "19/10/2014"

Taking the date as the input, we first split the data into an array (slt) to attain the date,month and year. We then use awk's mktime function to get a date spec that is then used by awk's sttftime function to get the date in the correct format.

With a csv, $0 would be the number corresponding to the comma delimited field that has the date.

The date command can format this for you:

date -d "2014-10-19" +%d-%b-%Y

which yields

19-Oct-2014

If you give sample lines of your file we can script it for you.

One way is using awk:

echo zzz,19/10/2014,aaa,bbb | awk -F, ' getline date; OFS=","; $2=date; print; close(cmd)'

which gives

zzz,19-Oct-2014,aaa,bbb

Assuming here that you wish to modifiy the 2nd field. Adjust "split($2,dateelem,"/");" in the awk script to change the field (column) number your date appears in

perl can do it, but it's not really for beginners. However, parsing dates is best handled by proper date handling libraries. perl has one in the box, Time::Piece

$ cat file
a,b,19/10/2014,d

$ perl -MTime::Piece -F, -lane '
 $F[2] = Time::Piece->strptime($F[2], "%d/%m/%Y")->strftime("%d-%b-%Y");
 print join ",", @F
' file
a,b,19-Oct-2014,d

I have done by below method

command

 j=`date -d "2014/10/19" +%b`;echo "19/10/2014" | awk -v j="$j" -F "/" 'print $1"-"j"-"$3'

Output

19-Oct-2014

Another awk approach. Uses basic awk functionality, so likely to work with non-GNU implementations.

$ cat infile
foo,19/10/2014,bar
foo2,19/01/2015,bar2
$ awk -F, 'print $1","substr($2,0,2)"-"substr("JanFebMarAprMayJunJulAugSepOctNovDec",substr($2,4,2)*3-2,3)"-"substr($2,7,4)","$3' infile
foo,19-Oct-2014,bar
foo2,19-Jan-2015,bar2
$