variable inside awk/sed in alias

Question

I'd like to write an alias that would print particular line from all files in a folder, i.e.:
$ print_line 20 - would print 20th line from all files in a folder.

I spent hour with various constructions in .bashrc (some examples below), with lot of different escaping, but so far nothing worked. Thanks for any help.

alias a="for f in create*; do sed -n '\$1{p;q}' $f; done"

alias b="for f in create*; do awk 'NR==\$1{print $0}' $f; done"

function l { r=$(echo \'"$1"{p;q}\'); sed -n "$r" *; }

Why? `awk 'FNR==20' *` would do it just as succinctly as `print_line 20 *` — Ed Morton, Mar 09 '14 at 01:59
Thanks for the comment, that'd do too. I'm not too familiar with awk and shell too, obviously. — Pida, Sep 16 '14 at 19:35

score 3 · Accepted Answer · answered Mar 08 '14 at 09:54

3

You can try this,

printline() { for f in create*; do sed -n "$1{p;q}" $f; done ;}

answered Mar 08 '14 at 09:54

sat

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score 0 · Answer 2 · edited May 23 '17 at 12:28

bash aliases, unlike those in csh, do not allow parameters; see Make a Bash alias that takes a parameter? and Passing argument to alias in bash, and How to pass command line arguments to a shell alias? and How do I include parameters in a bash alias?

So you need a function. If you have the GNU sed with the -s switch, the following will work:

   printline() { sed -ns "$1 p" create*;}

Otherwise, slightly modifying sat's nice answer, use a for-loop:

   printline() { for f in create*; do sed -n "$1 p" "$f"; done;}

Note the space after the { is necessary. The use of double-quotes in "$f" guards against filenames with bad characters, such as spaces or stars.

variable inside awk/sed in alias

2 Answers2