Pass shell variable as a /pattern/ to awk

Question

Having the following in one of my shell functions:

function _process () {
  awk -v l="$line" '
  BEGIN {p=0}
  /'"$1"'/ {p=1}
  END{ if(p) print l >> "outfile.txt" }
  '
}

, so when called as _process $arg, $arg gets passed as $1, and used as a search pattern. It works this way, because shell expands $1 in place of awk pattern! Also l can be used inside awk program, being declared with -v l="$line". All fine.

Is it possible in same manner give pattern to search as a variable?

Following will not work,

awk -v l="$line" -v search="$pattern" '
  BEGIN {p=0}
  /search/ {p=1}
  END{ if(p) print l >> "outfile.txt" }
  '

,as awk will not interpret /search/ as a variable, but instead literally.

Answer 1

Use awk's ~ operator, and you don't need to provide a literal regex on the right-hand side:

function _process () {
    awk -v l="$line" -v pattern="$1" '
        $0 ~ pattern {p=1; exit} 
        END {if(p) print l >> "outfile.txt"}
    '  
}

Here calling exit upon the first match as we don't need to read the rest. You don't even need awk, grep would be enough and likely more efficient and avoid the problem of awk's -v var='value' doing backslash processing:

function _process () {
    grep -qe "$1" && printf '%s\n' "$line"
}

Depending on the pattern, you may want grep -Eqe "$1"

Answer 2

awk  -v pattern="$1" '$0 ~ pattern'

Has an issue in that awk expands the ANSI C escape sequences (like \n for newline, \f for form feed, \\ for backslash and so on) in $1. So it becomes an issue if $1 contains backslash characters which is common in regular expressions (with GNU awk 4.2 or above, values that start with @/ and end in /, are also a problem). Another approach that doesn't suffer from that issue is to write it:

PATTERN=$1 awk '$0 ~ ENVIRON["PATTERN"]'

How bad it's going to be will depend on the awk implementation.

$ nawk -v 'a=\.' 'BEGIN {print a}'
.
$ mawk -v 'a=\.' 'BEGIN {print a}'
\.
$ busybox awk -v 'a=\.' 'BEGIN {print a}'
.
$ gawk -v 'a=\.' 'BEGIN {print a}'
gawk: warning: escape sequence `\.' treated as plain `.'
.
$ gawk5.0.1 -v 'a=@/foo/' BEGIN {print a}'
foo

All awks work the same for valid escape sequences though:

$ a='\\-\b' awk 'BEGIN {print ENVIRON["a"]}' | od -vtx1 -tc
0000000  5c  5c  2d  5c  62  0a
          \   \   -   \   b  \n
0000006

(content of $a passed as-is)

$ awk -v a='\\-\b' 'BEGIN {print a}' | od -vtx1 -tc
0000000  5c  2d  08  0a
          \   -  \b  \n
0000004

(\\ changed to \ and \b changed to a backspace character).

Answer 3

Try something like:

awk -v l="$line" -v search="$pattern" 'BEGIN {p=0}; { if ( match( $0, search )) {p=1}}; END{ if(p) print l >> "outfile.txt" }'

Answer 4

No, but you can simply interpolate the pattern into the double-quoted string you pass to awk:

awk -v l="$line" "BEGIN {p=0}; /$pattern/ {p=1}; END{ if(p) print l >> \"outfile.txt\" }"

Note that you now have to escape the double-quoted awk literal, but it is still the simplest way of accomplishing this.

Answer 5

You could use the eval function which resolves in this example the nets variable before the awk is run.

nets="searchtext"
eval "awk '/"${nets}"/'" file.txt