How to insert Command Line argument of shell script in AWK?

Question

I have to find all of the record which have a particular data which I am gonna pass as the command line argument.

awk expression is like this(Date is in this format :'02/08/2013')

cat records.txt| awk -F ',' '$4 ~ /02/08/2013/ {print $1 $2}'

Here 4th column is the date column.

What I want to do is that, provide the date as the first argument and compare it.

I tried this,But it is not working.

cat records.txt| awk -F ',' -v awkvar="$1" '$4 ~ /^"awkvar/ {print $1 $2}'

Here the date column starts with " quote, so I am telling to look for the records who start with "+awkvar the given date.

Can anyone help me with this?

Edit:

awk -F ',' -v var1="$1" '$4 ~ /^"2013/ {print $1 $2}' {This one is working, as I am directly comparing the record with 2013}

when I do this

awk -F ',' -v var1="$1" '$4 ~ /^"var1/ {print $1 $2}' , it does not return anything, what is the difference.

Gilles Quénot · Accepted Answer · 2013-02-14 11:43:19Z

4

To pass variables to awk, use

awk -v awkvar=$value '{print awkvar}'

That said, no need to pipe cat | awk (useless use of cat) so finally :

awk -F, -v awkvar="$1" '$4 ~ "^\""awkvar {print $1 $2}' records.txt

edited Feb 14, 2013 at 11:43

answered Feb 14, 2013 at 10:15

Gilles Quénot

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6 Comments

Sahil Over a year ago

yes that is what I did, I updated the post as well, Do you think it should work? I have been trying this since yesterday?

Sahil Over a year ago

when I do this cat data.csv| awk -F ',' -v var1="$1" '$4 ~ /^"2013/ {print $1 $2}' , I get the output BUT when I am doing this matching with first argument cat data.csv| awk -F ',' -v var1="$1" '$4 ~ /^"var1/ {print $1 $2}', It does not give me anything

Sahil Over a year ago

Many thanks It worked, Could you please explain what happened there?what happened to the slashes and what is the issue with double quotes?

Gilles Quénot Over a year ago

There's no real issue, just need to juggle with backslash and quotes. The slash is not mandatory and can be replaced by double quotes like in my snippet. It's more convenient for variable handling. For further informations : grymoire.com/Unix/Awk.html

Shashank Vivek Over a year ago

@sputnick:awk -v not working on my Unix.I tried 1st command "awk -v awkvar=$value '{print awkvar}'"

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