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Shell Programming and Scripting

Script help required!

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# 1  
Old 05-15-2005
Script help required!
Hi there,

i am trying to create a script that checks for the existence of users on the system, if they exist then their details should print on the screen with a message that the id is in use. I am having a bit of trouble with it.

Any ideas?

Cheers
Kev!
# 2  
Old 05-15-2005
for existing of users in the system ,
Code:
grep /etc/passwd  $user 
if test $? -eq 0
then
    echo "$user exsits"
fi

for existing login users in the system ,

Code:
who | grep $user
if test $? -eq 0
then
    echo "$user exsits"
fi


Note : Code not tested, but i'm sure it works.
# 3  
Old 05-15-2005
Thanks very much!

Kev
# 4  
Old 05-15-2005
Quote:
Originally Posted by bhargav
for existing of users in the system ,
Code:
grep /etc/passwd  $user 
if test $? -eq 0
then
    echo "$user exsits"
fi

Code not tested, but i'm sure it works.
Except for the lapse in concetration it would Smilie

Code:
grep $user /etc/passwd
if test $? -eq 0
then
    echo "$user exsits"
fi

# 5  
Old 05-15-2005
Quote:
Except for the lapse in concetration it would

Thanks for pointing this out reborg !
This is one of drwbacks of multi-therading of the mind.
# 6  
Old 05-22-2005
Hi again,

Still having problems with this script, it works to an extent but I think im missing something.

If i enter a number, say 232435 and the user exists, it should return a message saying the user id is in use, if not then it should display an appropriate message that it is in use.

What I have so far runs and displays the message but the id part of the script doesnt seem to be working.

Any ideas?
Kev
# 7  
Old 05-22-2005
Code:
#!/bin/ksh

echo "Which UID do you want to check?"
read UID

match=`grep "^[^:]*:[^:]*:$UID:.*$" /etc/passwd`

if [ -n "$match" ]; then
  echo "$UID is IN USE!"
  echo "$match"
else
  echo "UID $UID is free..."
fi

exit 0

Cheers
ZB
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