Bash – Use a shell variable with Awk

awkbashshell

I have this:

if [[ $1 = "-c" ]] && [[ $2 =~ ^[a-z A-Z]+$ ]]
then
awk -v var="$2" '{print $2}' something.txt
fi

What I want is with a comand like -c abc to show the lines in something.txt that include abc.

Best Answer

Use grep:

grep abc something.txt

Also note that using $2 in the shell refers to the second argument (as you seem to know), but in awk it is different. Your question seems to show a misunderstanding of this so I'll clarify it.

The shell requires the $ to refer to the value of a variable. So you refer to a variable named myvar by writing $myvar.

In awk to refer to a variable named myvar you just use its name—myvar. To refer to a literal string containing the letters m-y-v-a-r, you type "myvar".

The $ in awk is to refer to the field with a specific number. So $2 refers to the second field of the current line of the file. Or if you set myvar = "4", then $myvar refers to the fourth field of the file.

For just printing all lines of a file that match a given pattern, use grep—that's what it's designed for.

Related Solutions

Bash – awk + set variable inside the awk

Do you want something like this?

function get_column
{ 
    awk -v val=$1 '{print $val}' list.txt
}

Above is returning the column match with $1 passing to the function. if you really need print the line match with line number in $1 from function, instead use below.

function get_row
{ 
    awk -v val=$1 'NR==val{print ; exit}' list.txt
}

Or let shell evaluated and set the val value and print that within awk as following:

function get_column
{ 
    awk '{print $val}' val=$1 list.txt
}

function get_row
{ 
    awk 'NR==val{print ; exit}' val=$1 list.txt
}

Here you are passing val with only numbers and if val was contain backslash escape character you will encounter a problem which awk does C escape sequence processing on values passed via -v val= and a shell variable with val="\\n" will change to value with \n by awk.

Awk – How to Use Shell Variable Inside a Pattern

That should be:

if ($i ~ field_name "=[0-9]*.?[0-9]+") ...

Or:

 regexp = field_name "=[0-9]*.?[0-9]+"
 if ($i ~ regexp) ...

Note that . matches any single character. If you want to match a literal ., you'd need regexp to contain \. (which inside double quotes would have to be written \\.) or [.].

 regexp = field_name "=[0-9]*\\.?[0-9]+"

I'd also expect you'd want to anchor the regexp:

 regexp = "^" field_name "=[0-9]*\\.?[0-9]+$"

Other notes:

cat "${file}" is a UUOC which also has the drawback (over a redirection) that it doesn't work when $file starts with - and still runs awk if the file can't be opened.
-v field_name="$external_data" mangles backslashes. Another approach that doesn't have the problem is to use an environment variable: FIELD="$external_data" awk ... and refer to it within awk as ENVIRON["FIELD"].
as the contents of field_name is copied verbatim into regexp, it is treated as a regexp, so if $external_data contains regexp operators (.+*?{}()[]\^%...), it may not work properly.
in some locales and awk implementations, [0-9] matches a lot more characters than just 0123456789 (though I suspect it would be (non-ASCII) characters unlikely to occur in your input).

With perl:

FIELD=size <"$file" perl -lpe '
  s{
    (?<![^\t])       # not-preceded by a non-TAB
    \Q$ENV{FIELD}=\E # contents of $FIELD taken literally
    \K               # matched portion starts here
    \d*\.?\d+
    (?![^\t])        # not followed by a non-TAB
  }{
    sprintf "%.4f", $&
  }gxe'

Which would not have any of the issues discussed above (it would also work better than awk with input that contains invalid text, like a mix of text and binary data, or text encoded in a charset different from that of the user's locale).

Best Answer

Related Solutions

Bash – awk + set variable inside the awk

Awk – How to Use Shell Variable Inside a Pattern

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