How to use variable including special symbol in awk?

问题

For my case, if a certain pattern is found as the second field of one line in a file, then I need print the first two fields. And it should be able to handle case with special symbol like backslash.

My solution is first using sed to replace \ with \\, then pass the new variable to awk, then awk will parse \\ as \ then match the field 2.

escaped_str=$( echo "$pattern" | sed 's/\\/\\\\/g')    
input | awk -v awk_escaped_str="$escaped_str"  '$2==awk_escaped_str { $0=$1 " "     $2 " "}; { print } '

While this seems too complicated, and cannot handle various case.

Is there a better way which is more simpler and could cover all other special symbol?

回答1:

The way to pass a shell variable to awk without backslashes being interpreted is to pass it in the arg list instead of populating an awk variable outside of the script:

$ shellvar='a\tb'

$ awk -v awkvar="$shellvar" 'BEGIN{ printf "<%s>\n",awkvar }'
<a      b>

$ awk 'BEGIN{ awkvar=ARGV[1]; ARGV[1]=""; printf "<%s>\n",awkvar }' "$shellvar"
<a\tb>

and then you can search a file for it as a string using index() or ==:

$ cat file
a       b
a\tb

$ awk 'BEGIN{ awkvar=ARGV[1]; ARGV[1]="" } index($0,awkvar)' "$shellvar" file
a\tb

$ awk 'BEGIN{ awkvar=ARGV[1]; ARGV[1]="" } $0 == awkvar' "$shellvar" file
a\tb

You need to set ARGV[1]="" after populating the awk variable to avoid the shell variable value also being treated as a file name. Unlike any other way of passing in a variable, ALL characters used in a variable this way are treated literally with no "special" meaning.