Very Computer

My program generates a date output from which I need to extract the
output. It should be done such that year, month, day, hour and minute
fields should be extracted separately.

The date output is in this format.  2006/5/25-9:00. The month, day and
hour fields should have 2 decimals instead of 1.

I tried different ways in sed and awk to extract them but I am not
getting any results. Can anyone let me know how to do this.

Thanks,
doni

Try this one:

echo 2006/5/25-9:00 | \
awk 'gsub(/[/:-]/," "){printf "%i %.2f %.2f %.2f %i\n", $1, $2, $3, $4,
$5}'

Torfinn

Hi Torfinn,

I tried but it didnt work.
Here's the output that I got when I tried with ur method. "2006525900
0.000.000.00 0"

I believe you are almost there. Observe the single space in the gsub
replacement field and between format specifiers in the print format
string. See the following for hopefully better readability:

awk '
gsub(/[/:-]/,"              "){
printf "%i     %.2f      %.2f     %.2f     %i\n", $1, $2, $3, $4, $5

Quote:}      '

Tested using  gawk 3.1.5 .

echo 2006/5/25-9:00 | awk -F '[/:-]' '{print $1,$2,$3,$4,$5}'
2006 5 25 9 00

--

Thanks Torfinn and Michael. I tried both of your methods and they work
fine. I like Michael's way of doing the function.