printing fields using parameters

[kasparov]

Can someone tell me how I can use a variable as the parameter to a print startement? e.g.

VAR=7
echo $LINE | awk '{ print ??? }'

I want to print the 7th field in $LINE (but next time $VAR will have a different value). I've tried all the variations of dollar signs & brackets that I can think of but I either get the whole of $LINE or a syntax error.

TIA, Chris

 

[vgersh99]

VAR=7
echo $LINE | awk -v field="${VAR}" '{ print $field }' vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[kasparov]

Vlad

Thanks for the reply - unfortunately I'm still getting a syntax error:

awk: syntax error near line 1
awk: bailing out near line 1

I'm on Solaris 8 using ksh if that's any use?

Chris

 

[vgersh99]

use nawk instead of awk vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+