CODEWITHSUNDEEP
bashshellawk
Kostas Andrianos
Kostas Andrianos

Reputation: 1619

AWK: Passing two arguments and weird error

I have made an awk implementation of grep -c ^str file and I want to pass the file and str arguments from a shell script. I am using awk -v twice to pass the arguments but I get a awk: cannot open var1 (No such file or directory) error. I just can't get around it, I've been trying for almost an hour. My code:

read -p "Give me a file name " file
read -p "Give me a string " str
awk -v var1="$file" -v var2="$str" 'BEGIN{print var1; print var2}{/^var2/}' var1 | 
awk '{if ($0 != "/s") {count++}} END {print count}'

Upvotes: 1

Views: 98

Answers (1)

Juan Diego Godoy Robles
Juan Diego Godoy Robles

Reputation: 14955

It should be:

awk -v var1="$file" -v var2="$str" 'BEGIN{print var1; print var2}{/^var2/}' $file

awk vars can only be acceded inside awk code (delimited by single quotes in this case) not at shell level where var1 means nothing.

Note that var2 value will be just a literal string between slashes /^var2/, use $0 ~ "^"var instead to access var2 value.

In fact, your awk code can be rewritten as:

awk -v var="$str" '$0 ~ "^"var && $0 != "/s"{count++}END{print count}' $file

Upvotes: 2

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