Shell Programming and Scripting

Your requirements contradict each other.

Code:

print "Bad entries:\n\n";
while ($line = <FILEHANDLE>) {
    chomp($line);
    @fields = split /:/, $line;
    if ($fields[4] != 90) {
        print "$fields[0]\n";
    }
}

Hi Bala,

Thanks for your input:
As per your code, the output is like :

But, the output what expected is as below:

else it should be declare as GOOD. Hope it makes sense.

@gsiva: I'm still not able to understand your very specific (and contradicting) requirements.

And by now from all previous posts, you sure do have enough tools equipped to make minor modifications to the code yourself, to suit your requirement. Go ahead, give it a try and let us know what you have brewed :-)

The message does not tell you what the 5th value is exactly. We know that it is not 90, but is it 89 or 91 or 100 or something else? Maybe you want to see it; maybe you don't. But I'll assume that you do.

The message "GOOD" does not tell you what the ID was in the line you read. You'd simply see the text "GOOD" per line if all lines satisfy the condition. Maybe you want to see the ID; maybe you don't. But I'll assume that you do.

Given these criteria, a short script could be like this:

Code:

awk -F: '{
           if ($5==90) {
             print "GOOD: ID="$1,"has the 5th value 90 in /etc/shadow"
           } else {
             print "BAD: ID="$1,"does not have the 5th value 90 in /etc/shadow. It has",$5,"instead."
           }
         }' /etc/shadow

- Modify the print function in the "else" branch if you don't want to view the bad value.
- Modify the print function in the "if" branch if you only want to see "GOOD"
- Remove the "else" branch completely if you only want to test for and print good lines.
- Remove the "if" branch completely and change the "else" branch to a converse "if" branch if you only want to test for and print bad lines.

tyler_durden