5

I'm trying to write a simple bash script in which the user inputs their username, then they are greeted, depending on the time of day by their surname. I currently have the following:

echo Please enter your username
read username
name=$(grep $username /etc/passwd | cut -d ':' -f 5)

h='date +%H'

if [ $h -lt 12]; then
  echo Good morning ${name::-3)

etc. etc.

I have managed to cut the 3 commas off the end of the name that are there, but I want to be able to cut the first name off.

For example:

  • The $name is Amber Martin,,,.
  • I've cut down to Amber Martin.
  • I need to cut down further to Martin.
  • And this needs to work with any name.
  • 1
    Did you mean h=$(date +%H) rather than h='date +%H'? – Toby Speight Apr 27 '17 at 14:10
  • 3
    I'm guessing this is just for an exercise or a personal thing, since parsing names and getting everything correct for every edge case is basically impossible. See kalzumeus.com/2010/06/17/… – Muzer Apr 27 '17 at 14:57
  • 3
    You don't. Guessing names is incredibly complicated. For example, here are some people all of whose last name is "Smith": John Smith, J. W. Smith, John William Smith, John-Jack Smith, John William Watson Smith. Basically, there are so many different naming conventions in different places and cultures that doing this 100% correctly is exceedingly hard. – terdon Apr 27 '17 at 16:31
  • Plus, even if you decide that surname == last name, you need to decide what happens if Cher or Madonna sign up. – origimbo Apr 27 '17 at 17:23
  • 1
    ah the old, Juan Pablo Fernández de Calderón García-Iglesias problem – Neil McGuigan Apr 27 '17 at 19:50
11

Better to use getent passwd than to read /etc/passwd directly. getent also works with LDAP, NIS and such. I think it exists in most Unixes. (My OS X doesn't have it, but it doesn't have my account in /etc/passwd either, so...)

name=$(getent -- passwd "$USER" | cut -d: -f5)

The string processing can be done with the shell's parameter expansion, these are POSIX compatible:

name=${name%%,*}         # remove anything after the first comma
name=${name%,,,}         # or remove just a literal trailing ",,,"
name=${name##* }         # remove from start until the last space
echo "hello $name"

Use ${name#* } to remove until the first space. (Just hope no-one has a two-part last name, with space in between).

The cut could also be replaced with word-splitting or read, by setting IFS to a colon.

  • This will only work if the unwanted part is literally ,,, (i.e. no location or phone information). It would be more robust to cut everything from the first comma: name=${name%%,*}. – Toby Speight Apr 27 '17 at 14:03
  • Is this compatible with every *nix? Or only linux? – Luciano Andress Martini Apr 27 '17 at 14:08
  • @TobySpeight, ah, yes, that's true. (Hopefully no-one's name contains a comma) – ilkkachu Apr 27 '17 at 16:29
  • 2
    @LucianoAndressMartini, the prefix and suffix -removing string operations are from POSIX, I think getent is common. – ilkkachu Apr 27 '17 at 16:34
5
#!/bin/bash
#And also /bin/sh looks like to be compatible in debian.  
echo "Hmmm... Your username looks like to be $USER"
name="$(getent passwd $USER | cut -d: -f5 | cut -d, -f1)"
echo "Your full name is $name"
surname="$(echo $name | rev | cut -d' ' -f1 | rev)"
echo "Your surname is $surname"
echo "thank your for using only cut and rev to do that..."
echo "But i am sure there is a better way"
4

Once you have the GECOS (comment) field, you can simply perform another cut to remove the (empty in your case) location and phone number fields, this time with , as the separator:

name=$(getent passwd "$USER" | cut -d: -f5 | cut -d, -f1)
echo "Hello, ${name##* }-san!"

I'll leave it as an exercise to deal with all the different possibilities of what is a "surname"!

  • 1
    Upvoted for the "falsehoods" link. :) – shoover Apr 27 '17 at 16:01

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