I need to print the ASCII value of the given charecter in awk only.
the below gives 0 as output.
echo a | awk '{ printf("%d \n",$1); }'
Help please.
views:
174answers:
3
A:
It seems this is not a trivial problem. I found this approach using a lookup array, which should work for A-Z at least:
BEGIN { convert="ABCDEFGHIJKLMNOPQRSTUVWXYZ" }
{ num=index(convert,substr($0,2,1))+64; print num }
Martin Wickman
2010-03-23 11:17:27
for this solution to work i need to create lookup array for all the characters in the ASCII table in the order.
Venkataramesh Kommoju
2010-03-23 11:27:45
Yes, that is correct. It's only 127 values though, so it's not that big a deal.
Martin Wickman
2010-03-23 12:08:53
+3
A:
see the awk manual for ordinal functions you can use. But since you are using awk, you should be on some version of shell, eg bash. so why not use the shell?
$ printf "%d" "'a"
97
ghostdog74
2010-03-23 11:21:42