I have a file having multiple lines in the following order
Name=abc Date=12/10/2013 Name=xyz Date=11/01/2014 Name=pqr Date=06/30/2014 Name=klm Date=07/08/2014 And so on. Date format is mm/dd/yyyy.
I want to write a shell script, which will traverse each line and it should return me the line, whose Date is about to come in next 1 week. Like here, if I run the script today, it must return me
Name=pqr Date=06/30/2014 How to achieve that ?
# Get today's date in natively sortable format: YYYYMMDD today=$(date +%Y%m%d) # Get next week's date in natively sortable format: YYYYMMDD oneweek=$(date -d "+1 week" +%Y%m%d) # Pass the start and end dates to awk as variables. # Use `=` as the awk delimiter. # Split the third field on `/` to convert the date to the natively sortable format. # Compare converted date to start and end values (as an `awk` pattern so true results use the default `awk` print action). awk -v start=$today -v end=$oneweek -F= '{split($3, a, /\//)} (a[3] a[1] a[2] > start) && (a[3] a[1] a[2] < end)' file 
oneweek=$(date -d "+1 week" +%Y%m%d)...... it showed me a warning date: illegal option -- d...... :(
GNU coreutils date. It would seem the AIX date does not have that flag. Perhaps it has the long --date option instead?--date option also, but it didn't work..... I was not able to see the version of date also in my AIX by date --version...... Though all working good in Linux machineSimple way is get date from each line and convert it to seconds from Epoch time and compare as shown below.
while read line
do
date_in_line=`echo $line | awk -F= '{print $NF}'` date_after_1_week_in_sec=$(date -d "+1 week" +%s) date_in_line_sec=$(date -d $date_in_line +%s) if [ $date_in_line_sec -lt $date_after_1_week_in_sec ]; then echo $line fi done < file.txt
Hope it helps.
datecommand you can output a time using relative offsets likedate --date="+1 week".