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I am trying to use awk inside a shell script function, but can't seem to find a reference that tells me how to properly escape the awk variable ($4). Inside the function it keeps getting replaced at the 4th item passed to the function (Null).


iHealthGetFile()
{
    fileID=$(cat qkViews/${dir}iHealth-Files.json|grep -B5 '${1}'|grep id|awk '{print $4}'|sed 's/,//g; s/\"//g')
    echo $fileID
    echo "Retrieving file $1 as $fileID from ID: $ID"
}

(disclaimer: Yes, I could probably do something differently with the piped commands... but my main goal is to figure out the escaping of the $4 first.)

  • 3
    awk '{print $4}' is the correct way to return the fourth field. You have put $4 in single-quotes which is the correct way to protect it from the shell. The error is caused by something else. Provide details and maybe we can help. – John1024 Dec 10 '17 at 22:58
  • @john1024, changing $4 in the awk script portion to '$4' still doesn't prevent the $4 from being evaluated as a parameter to the function, unfortunately. – Jerry Lees Dec 11 '17 at 3:45
  • The way that you have it in the question, awk '{print $4}', is correct if you are expecting awk to evaluate $4 as the fourth field on the input line. The single quotes prevent the shell from altering the command before awk is run. – John1024 Dec 11 '17 at 3:51
  • Did @roaima's answer work for? If not, you need to make more clear what your input looks like and what output you expect. – John1024 Dec 11 '17 at 3:52
3

As far as I can tell without any contextual data, the error is not in your awk code, but within your first grep, which is looking for the literal four characters ${1}. I suspect this was intended to be the first parameter passed to your function.

I've altered your code a little. If you really intended the echo $fileID to remove extraneous whitespace you'll want to remove the double quotes I've added around the variable there.

You're also using a couple of global variables. Mixing local variables ($1) and global variables ($dir and $ID) is not particularly good coding practice, but there are times in a script where pragamatism may need to win.

iHealthGetFile()
{
    fileID=$(
        grep -B5 "$1" "qkViews/${dir}iHealth-Files.json" |
        awk '/id/ {print $4}'|
        tr -d ',"'
    )
    echo "$fileID"
    echo "Retrieving file $1 as $fileID from ID: $ID"
}
  • The $1 is correct, it's the filename passed to the function. The function is called like so, "iHealthGetFile /some/directory/some/FileName" In this case, like any shell script function, $1 represents the first parameter passed when the function is called. This is desirable. However, the replacement of $4 in the awk command is not since it is not intended to be the fourth parameter passed, but instead part of the awk function. – Jerry Lees Dec 11 '17 at 0:09
  • 1
    @JerryLees great; I've understood what your code is trying to do. The problem was in the first grep statement. There is nothing wrong with your awk statement as presented in your question – roaima Dec 11 '17 at 8:24
0

In your question you claim that awk returns null as a result of $4 passed to the shell function is null.

The reason that your awk (or your whole pipe) returns null is not due to a kind of confussion with shell function parameters.
It is just because whatever is feeding awk with data (left part of your awk pipe) has not a valid 4th field .

This test will prove it:

$ function tt { echo "1:$1-2:$2-3:$3-4:$4"; echo "first second third fourth" |awk '{print $4}'; }
$ tt one two #call the function with only two parameters. 3 and 4 are null
1:one-2:two-3:-4: #Parameters passed to function
fourth #awk result

As you can see , awk result is not affected by function parameters . If the data provided to awk are valid, 4th field of these data will be returned.

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