Need some help with a regex if loop problem

Shell Programming and Scripting
1

Need some help with a regex if loop problem.
File1:

2323
3232
4230
3230
4340
4343
233
32320

I want to print "Zero" if the number ends with a zero, but print "number" if it does not!

#!/bin/bash
/usr/bin/nawk '{
if ($1 ==/[0-9]+0\b/){
 print "Zero"}
else{
 print "number"}
}' File1

this code just prints out "Number" for each line! Help!

2

One way:

awk '{print $0%10?"Number":"Zero"}' file
3 
awk '{if($0 ~ "0$") {print "Zero"} else {print "number"}}' yourfile

edit: one more:

sed "s/.*[1-9]$/number/g;s/.*0$/zero/g" yourfile
4

The "~" operator is for matching regular expressions.

$
$ awk '{if ($0 ~ /0$/) {print $0,"Zero"} else {print $0,"number"}}' f1
2323 number
3232 number
4230 Zero
3230 Zero
4340 Zero
4343 number
233 number
32320 Zero
$
$ # or alternatively -
$ awk '/0$/ {print $0,"Zero"} !/0$/ {print $0,"number"}' f1
2323 number
3232 number
4230 Zero
3230 Zero
4340 Zero
4343 number
233 number
32320 Zero
$
$

tyler_durden

5 
while read -r LINE
do
  case "$LINE" in
    *0 ) echo "zero";;
    *[1-9]) echo "number";;
  esac
done <"file"
6

In addition to the operator error pointed out by durden_tyler, \b in an awk regular expression is an escape sequence which matches the backspace character (unlike in perl where it refers to a word boundary).