Bug in shell script using awk to read /etc/passwd?

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.

Dave's Answer:

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. :-)

Categorized: Shell Script Programming (Article 7289, Written by Dave Taylor)
Tagged: awk, login, scripting, shell script
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