Shell Programming and Scripting

Hi
i need to convert a date in the format DD-Mon-YY to YYYYDDMM

Ex : 01-JUL-00 to 20000701

Can anybdy help me with this??

Thanks in advance
Shenaz

Hi shanu using '-' as delimiter u can awk but i am not sure how can we directly convert "JUL" to 07. might be you need to place all the months and their numbers in a file first and awk.

An example how to convert the date format DD-MMM-YY to YYYYMMDD assuming the century is 2000 if the year is in 2 digits:

Code:

awk -v var="01-JUL-00" '
BEGIN{
  split("JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC", month, " ")
  for (i=1; i<=12; i++) mdigit[month[i]]=i
  m=toupper(substr(var,4,3))
  dat="20"substr(var,8,20)substr(var,1,2) sprintf("%02d",mdigit[m])
  print dat
}'

Regards

awk -v does nt seem to work.

awk -v var="01-JUL-00" '
BEGIN{
split("JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC", month, " ")
for (i=1; i<=12; i++) mdigit[month[i]]=i
m=toupper(substr(var,4,3))
dat="20"substr(var,8,20)substr(var,1,2) sprintf("%02d",mdigit[m])
print dat
}'
awk: syntax error near line 1
awk: bailing out near line 1

So i tried this

echo $var | awk 'BEGIN{FS="-";OFS="";split("JAN FEB MAR APR MAY JUN JUL AUG SEP OCT NOV DEC", month, " ")
for (i=1; i<=12; i++) mdigit[month[i]]=i
m=toupper(substr(var,4,3))
dat="20"substr(var,8,20)substr(var,1,2) sprintf("%02d",mdigit[m])
print dat}'

and the output is 2000 and not 20070701

Any suggestions??

Use nawk, gawk or /usr/xpg4/bin/awk on Solaris.

Regards

Great!!!! It works!!!

Thanks