Using variables to initialize regular expressions in awk - variables

Question

I want to initialize a variable with a regular expression, and then use it for pattern matching. Results do not come as expected . So for example I have,
BEGIN {
item_code_pattern=/ITM-CD-10/ ;
}
$0 ~ $item_code_pattern{ print ; }
I see that records which do not have pattern as ITM-CD-10 are also coming in the output.
Please suggest what should be the correct boolean expression before the block.
Thanks

Answer 1

You want to use a regular string:
awk '
BEGIN {
item_code_pattern = "ITM-CD-10" ;
}
$0 ~ $item_code_pattern { print ; }
'
The /pattern/ construct checks whether $0 matches the given pattern, so your original code is equivalent to saying:
item_code_pattern = $0 ~ "ITM-CD-10"
Since $0 is empty in the BEGIN section, item_code_pattern is set to 0.

Answer 2

You need to drop the $ and the / symbols (and there's no need for a BEGIN block, just assign the variable on the command line):
awk '$0 ~ item_code_pattern' item_code_pattern=ITM-CD-10
When you use $, some versions of awk will emit an error while others will silently convert the variable to an integer value of 0 so that $item_code_pattern is exactly the same as $0, and the code $0 ~ $item_code_pattern is the tautology $0 ~ $0.
If you insist on using a BEGIN block, the syntax is:
BEGIN { item_code_pattern="ITM-CD-10" }
$0 ~ item_code_pattern
Note that { print } is the default rule when no rule is given, so it is redundant.