awk: How pass parameters to the function, called from inside awk ?

[hellish]

awk: How pass parameters to the function, called from inside awk ?

Hello, Experts!

Please, help me to figure this out:

Here is sample function:

$ testfn() { echo "${1} world"; }
$ testfn hello
hello world

awk example:

$ echo "something bla bla2"|awk '$1 ~/something/ { print $1; print $0; }'
something
something bla bla2

Now, I want to change "something" to "something world" using created above function, when printing it as a whole line, by passing first awk "column element" as a parameter:

$ echo "something bla bla2"|awk '$1 ~/something/ { $1="'"$(testfn) $1"'" ; print $0 }'
 world  bla bla2

^^Above doesn't work^^

$ echo "something bla bla2"|awk '$1 ~/something/ { $1="'"$(testfn)"$1'" ; print $0 }'
 world bla bla2

^^Above doesn't work^^

$ echo "something bla bla2"|awk '$1 ~/something/ { $1="'"$(testfn)"'$1" ; print $0 }'
 world$1 bla bla2

^^Above doesn't work^^

Is thee any way to pass parameters from awk to the function inside awk ?


Thanks!

[watael]

hi,

why don't you declare the function inside awk ?

awk 'function sayHello(arg)
{
print "Hello "arg
}
/^something$/{sayHello($1)}'

[hellish]

Thanks! And would it be possible to call shell function from declared function inside awk ?