Bug in Shell Script Using awk to Read /etc/passwd?

Dave Taylor By Dave Taylor
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In your book Wicked Cool Shell Scripts, on p. 115, Script #40 Reporting Disk Hogs, on the 8th line the second part of the line reads:

awk -F: '$2 > 99 {print $1} ')

I am very new to shell scripting, so I am probably just not understanding how this line should be interpreted. In my Linux class, I have been learning that in the /etc/passwed file the 1st field is username, the 2nd field is password (x), and the third field is the UID, so wouldn’t the $2 above really be $3 so that the script would be checking to see if the UID field, which is field 3 is greater that 99, not the password field,which is field 2? Please let me know, since this is confusing to me. Thanks for your help.

You’re right. But you haven’t noticed that this is part of a pipe in that particular script:

for name in $(cut -d: -f1,3 /etc/passwd | awk -F: '$2 > 99 { print $1 }')

The cut just extracts the name and uid from the password file, at which point the second field *is* the UID, and this way we can use awk to only print the username of accounts with uid’s greater than 99.

Remember, here’s a typical few lines from the Linux / Unix /etc/passwd file:

coloradoport:*:1041:1041:coloradoportraits:/
home/coloradoport:/sbin/nologin
shining-light:*:1042:1042:shining-light-books:/home/shining-light:/sbin/nologin
startup101:*:1043:1043:startup101:/home/startup101:/sbin/
nologin

Of course, looking at it now, I don’t know why I didn’t just use:

for name in $(awk -F: '$3 > 99 { print $1 }' /etc/passwd)

But that’s another story. :-)

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About Dave Taylor
Dave Taylor is known as an expert on both business and technology issues. Holder of an MSEd and MBA, author of twenty books and founder of four startups, he also runs a marketing company and consults with firms seeking the best approach to working with weblogs and social networks. Dave is an award-winning speaker and frequent guest on radio and podcast programs.

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