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trim 0 ahead of a time,pls help~

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# 1  
Old 07-01-2010
trim 0 ahead of a time,pls help~
Hi，

I am trying to write a ksh to compare the time in a date

Code:
date
Thu Jul  1 09:01:24 PDT 2010


when I try to get hour

Code:
date | awk '{print $4}' | cut -f1 -d:
08

how I can trim the 0 ahead of 08 to make it 8?

please help~
Last edited by Scott; 07-01-2010 at 02:23 PM.. Reason: Code tags
# 2  
Old 07-01-2010
date +%l (lower case L)
# 3  
Old 07-01-2010
The date command can output just the hour field without the leading zero.

Code:
date "+%1.1H"

# 4  
Old 07-01-2010
If your Korn shell is ksh93:
Code:
printf "%(%-H))T" now
12
printf "%(%-H))T" "4 hours ago"
8

# 5  
Old 07-01-2010
what about if I want to get the hour in ls -lrt?

Code:
abc*abc*/u01/app/abc/local/batch_log>ls -lrt
-rw-r--r--    1 oracle dba          348236 Jul 01 09:15 mon2.sh_100701.log


ls -lrt  mon2*.log | tail -1 |awk '{print $8}' | cut -f1 -d:

09

I need to show as 9 instead of 09
Last edited by Scott; 07-01-2010 at 02:24 PM.. Reason: Code tags
# 6  
Old 07-01-2010
Code:
ls -lrt  mon2*.log | tail -1 | awk '{print $8}' | cut -f1 -d: | sed 's/^0//'

# 7  
Old 07-01-2010
Code:
ls -lrt mon2*.log | tail -1 | awk -F"[ :]" '{print int($8)}'
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