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Using variable inside awk

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# 1  
Old 08-06-2009
Using variable inside awk
I am trying to print the lines with pattern and my pattern is set to a variable express

Code:
awk '/$express/{where=NR;print}' test2.log

I am not getting any data even though i have the data with the pattern. Can seomeone correct me with the awk command above?
# 2  
Old 08-06-2009
Use " instead of ' or awk -v. There is a search function in this forum - this is asked quite often, also for sed.
# 3  
Old 08-06-2009
I am using awk -v in a bash file. unable to get the result. Will it be teh same in a bash file?
# 4  
Old 08-06-2009
Did you try using " as zaxxon said ???
ex:
Code:
awk "/$express/{where=NR;print}" test2.log

# 5  
Old 08-06-2009
My total code is like this. I am using this bash file to get the data that is 5 hrs before the current time.

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 $express
awk "/$express/{print}" test.log

the statement is bailing out here. where as it is working fine if i pass a string to the variable like express='Parameters'

the output is coming like this:
Code:
08/06/2009 07
awk: syntax error near line 1
awk: bailing out near line 1

# 6  
Old 08-06-2009
Before putting into awk , check whether the variable contain any thing (i mean non space ) , some thing like this :
Code:
if [ "$variable" = " " ]; then
#  variable is space 
else
awk ...........
fi

# 7  
Old 08-06-2009
i am getting the error as follows. I am able to echo the result of the variable express correctly.
Code:
08/06/2009 07
test.sh: line 22: syntax error near unexpected token `else'
test.sh: line 22: `else'
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