How can I differentiate between $1 as a shell variable and $1 as an awk variable?

Question

When I pass the Command Line arguments to my script, We have to access that value by using $1, $2 etc., Similarly, when we are using awk, we can mention the columns of a file as $1,$2 etc.,

 #!/bin/bash
 cat $1 | awk '$1 > 20'

Let us assume I have passed a file name as a command line argument, and we can refer that argument by using $1.

Let us assume the file content is as follows.

How can I tell, when reading this code, whether it will access the command line $1 -- the input file name -- or the first column of the output from cat?

They are two different `$1`, one a shell positional argument number 1 and the other `$1` is AWK first field. — Diego Torres Milano, Nov 22 '17 at 19:51
@DiegoTorresMilano yeah, they are different, But how can I differentiate while I was writing the script and using awk in that same script. — Naga Venkatesh Gavini, Nov 22 '17 at 19:52
I've edited the question's title to better reflect the question as clarified in your above comment. — Charles Duffy, Nov 22 '17 at 20:13
BTW, `cat $1` is generally buggy. It won't work with a file named `my inputs.txt`, for example, or a file named `-inputs.txt` unless its name is preceded with `./` or other elements. If you *must* use `cat`, use it as `cat -- "$1"` -- but it's better to just pass the filename straight to `awk`, or to use `<"$1"` to perform a shell redirection. — Charles Duffy, Nov 22 '17 at 20:15

score 2 · Accepted Answer · answered Nov 22 '17 at 20:11

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There's an easy rule here: The shell does not perform any expansions on single-quoted strings.

This is true in other scenarios as well: echo '$1' prints the exact string $1.

Thus, in the following code (edited for correctness -- lack of double quotes around the shell variable caused a bug, and use of cat had needless inefficiency):

awk '$1 > 20' <"$1"

...we have $1 > 20 treated as literal by the shell and parsed as code only by awk, and "$1" -- being in double quotes, expanded by the shell.

answered Nov 22 '17 at 20:11

Charles Duffy

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So if the first argument to the script is "myfile.txt", this runs the awk script `$1 > 20`with input from myfile.txt. Conversely, if you swap the double- and single-quotes to `awk "$1 > 20" <'$1'`, it would run the awk script `myfile.txt > 20` (which doesn't make any sense), and try to read input from a file literally named "$1" (which doesn't make much sense either). Basically, it's all determined by how things are quoted. – Gordon Davisson Nov 22 '17 at 23:02
@charles Duffy, Now If I want to take 20 as the Second command line Argument, Now I have to make my code as this [awk '$1 > "$2" ' < "$1"], But the code is displaying entire Data. Any Help !! – Naga Venkatesh Gavini Nov 23 '17 at 04:44
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That's almost certainly incorrect quoting, but we can't guess what you are actually trying to say. Perhaps something like `awk -v limit="$2" '$1 > limit' <"$1"` (though really the redirection is superfluous here; like most file-oriented commands, Awk accepts a list of input files to read). – tripleee Nov 23 '17 at 06:39
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@NagaVenkateshGavini, if that's your problem, there's a separate question for it: [How do I use shell variables in an awk script?](https://stackoverflow.com/questions/19075671/how-do-i-use-shell-variables-in-an-awk-script) – Charles Duffy Nov 23 '17 at 15:38

score 0 · Answer 2 · answered Nov 22 '17 at 19:35

Is there is any another way to represent the columns while using awk in the shell script?

You can represent them using variables For example

awk '{for(i=1;i<NF;i++){print $i}}' file

Here NF is the total number of fields(which may be considered as columns.) As i gets different values in loop the $i would expand to individual columns like $1, $2 and so.

How can I differentiate between $1 as a shell variable and $1 as an awk variable?

2 Answers2