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Handling regular expressions in awk

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# 1  
Old 12-02-2009
Handling regular expressions in awk
Script is: accept filename as argument(also handle CTRL+C).to check whether th file exist in the current directory,it it then using awk find the employees who are either born in 1990 or drawing a salary greater than 25000.
In my database file 1st field is of id ,2nd field is name,5th field is of salary and 4th field is of date of birth in the format dd/mm/yyyy.

Error is: Unexpected end of file and unable to match pattern "'.

Please check my code and also suggest me how to handle CTRL+C



code :
Code:
if test -e $1
then
        echo "File Exist" in the current directory"
fi
echo "Employee details born in 1990 or drawing a salary greater than 25000"
awk -F"|" '$5 >25000 || $4 ~/90$/
{
        printf "%d %s %s %d\n", $1,$2,$4,$5
}' emp.lst


Thanks in advance.
Last edited by pludi; 12-02-2009 at 04:35 AM.. Reason: code tags, please...
# 2  
Old 12-02-2009
You do have an extra quote in this line:

Code:

echo "File Exist" in the current directory"

And you should probably quote $1 here:
Code:

if test -e "$1"

What is the purpose of handling Control-C?

Code:
trap "echo do something here...." 2

Last edited by Scott; 12-02-2009 at 05:12 AM..
# 3  
Old 12-02-2009
problem in awk scripts
Thanks..... that quote problem is solved
but the script written in awk is not printing the output.....only printing the correct value for 1st field rest 0 is printing.My databas file is emp.lst and i have put a space between > and 25000 then also not working.....please help me out.

i run this script as: sh empawk.awk emp.lst

empawk.awk is my code file.
# 4  
Old 12-02-2009
The only problem with your awk that I can see is that the { should not be on a newline:

Code:
awk -F"|" '$5 > 25000 || $4 ~/90$/ {
        printf "%d %s %s %s\n", $1,$2,$4,$5
}

Can you show a sample of your input? Is it really pipe-separated?

(also, if you have nawk, try that too, instead of awk!)
# 5  
Old 12-02-2009
Priyanka,

I tried this and it is working fine,

Code:
if test -e $1
then
        echo "File Exist in the current directory"
fi
echo "Employee details born in 1990 or drawing a salary greater than 25000"
awk -F"|" '$3 > 25000 || $4 ~/90$/ \
{
        printf "%d %s %s %d\n", $1,$2,$4,$5
}' emp.lst

Last edited by Neo; 12-02-2009 at 08:09 AM.. Reason: please use code tags!
# 6  
Old 12-02-2009
handling regular expression in awk
Thanks a lot scottn
u were right..... my input is not pipe separated.i made the fields pipe separated..then the same code is working fine........
thanks a lot again....
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