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Using variable inside awk

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# 8  
Old 08-06-2009
please post your script so we can better help you
# 9  
Old 08-06-2009
Can You Try something like this :

Code:
if [ "$express" = " " ]; then
echo "space"
else
awk -v ex="$express"  '/ex/ {print ex}'  rem.txt
fi

If wont work , Please post your complete script and the output your expecting.
# 10  
Old 08-06-2009
My basic requirement is to grep for the previous hour data. and my timestamp format is something like thisshown below.
Code:
(08/06/2009 05:48:45.992)(:)  
(08/06/2009 05:48:46.641)(:)

as there is a zero before the Hr, i can't just use
Code:
`date '+ %m/%d/%Y H'`

Instead i am using a bash file as shown below. I am able to get the date format but while using awk i am getting error. Please help me if there is any alternative for this requirement or any change needed in awk format? I am using sunOS box.
Code:
#!/bin/bash
genexpression()
{
stime=`date '+ %H'`
stime=`expr $stime \- 1`
if (( $stime < 10 ))
then
stime=0${stime}
fi
export express=`date '+%m/%d/%Y'`\ $stime
}
genexpression
echo $express
awk "/$express/{print}" test.log

# 11  
Old 08-06-2009
im not sure what you meant by "you can't use". Maybe you missed the % sign
Code:
date +"%m/%d/%Y %H"

# 12  
Old 08-06-2009
I did use the % but the grep is failing because it identifies hrs as teh file name and error out as ' unable to open 06' if 06 is the hrs.

---------- Post updated at 09:26 AM ---------- Previous update was at 09:23 AM ----------

adding to that i need the previous hr data. suppose if the current time is 08/06/2009 09:00
I need the data from 08/06/2009 08:00 to 08/06/2009 08:59
# 13  
Old 08-06-2009
some thing like this :

Code:
#!/bin/bash
genexpression()
{
stime=`date '+ %H'`
stime=`expr $stime \- 5`
if (( $stime < 10 ))
then
stime=0${stime}
fi
export express=`date '+%m/%d/%Y'`\ $stime
}
genexpression

echo $express

t=`grep -w "$express" log.txt | wc -l`

if [ $t -ne 0 ];then
echo "suces"
else
echo "fail"
fi

# 14  
Old 08-06-2009
See my Four Ways to Pass Shell Variables in AWK
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