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Frustrating Groups loop

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# 1  
Old 07-10-2012
Frustrating Groups loop
I'm trying to list a bunch of users' groups and filter them out to be space delimited like this:
Code:
3610 14
25 2810

Using the "id" command they would look like this:
Code:
uid=39693(user1) gid=14(dev) groups=3610(a_dev),14(dev)
uid-39471(user2) gid=25(testing) groups=2810(prod),25(testing)

I want the script to echo a blank line if the "id" command returns no such user. However, my condition doesn't appear to like my syntax and I've viewed several other links about syntax inside the double brackets and the script will not accept my condition!
Code:
while read i
do
USER_HAS_GROUPS="`id $i`"
USER_LDAP_GROUPS="`id $i | cut -d\= -f4 | sed 's/([^()]*)//g;s/,/ /g'`"
if [[ "$USER_HAS_GROUPS" == *such* ]]
then
echo ""
else
echo $USER_LDAP_GROUPS
fi
done < txt1

I've tried variants such as without the quotes, and quotes around *such*, single brackets, single equals sign, etc. Doesn't that condition mean "if $USER_HAS_GROUPS contains the string "such" anywhere then the condition is true" ?
# 2  
Old 07-11-2012
The "no such user" message is probably being output to stderr to capture it you will need to redirect stderr to stdout like this:
Code:
USER_HAS_GROUPS="`id $i 2>&1`"

Here is another thought, why not examine the return value from the id command something along the lines of this:
Code:
while read i
  if id $i > /dev/null 2>&1
  then
    USER_LDAP_GROUPS="`id $i | cut -d\= -f4 | sed 's/([^()]*)//g;s/,/ /g'`"
    echo $USER_LDAP_GROUPS
  else
     echo "Invalid user $i"
  fi
done < txt1

# 3  
Old 07-11-2012
Quote:
Originally Posted by Chubler_XL
The "no such user" message is probably being output to stderr to capture it you will need to redirect stderr to stdout like this:
Code:
USER_HAS_GROUPS="`id $i 2>&1`"

Here is another thought, why not examine the return value from the id command something along the lines of this:
Code:
while read i
  if id $i > /dev/null 2>&1
  then
    USER_LDAP_GROUPS="`id $i | cut -d\= -f4 | sed 's/([^()]*)//g;s/,/ /g'`"
    echo $USER_LDAP_GROUPS
  else
     echo "Invalid user $i"
  fi
done < txt1

Not only does that kludge not work for me but I don't understand why...it keeps saying "-bash: syntax error near unexpected token `done'". I believe the syntax is correct...there's a while followed by a done, I even put in a "do" just in case. The if statement is properly enclosed with fi...not sure why this is happening but I can't verify your suggestion.
# 4  
Old 07-11-2012
Code:
while read i
do
  if id $i > /dev/null 2>&1
  then
    USER_LDAP_GROUPS="`id $i | cut -d\= -f4 | sed 's/([^()]*)//g;s/,/ /g'`"
    echo $USER_LDAP_GROUPS
  else
     echo "Invalid user $i"
  fi
done < txt1

simple syntax error - missing "do"
This User Gave Thanks to jim mcnamara For This Post:
MaindotC
# 5  
Old 07-11-2012
Quote:
Originally Posted by jim mcnamara
Code:
while read i
do
  if id $i > /dev/null 2>&1
  then
    USER_LDAP_GROUPS="`id $i | cut -d\= -f4 | sed 's/([^()]*)//g;s/,/ /g'`"
    echo $USER_LDAP_GROUPS
  else
     echo "Invalid user $i"
  fi
done < txt1

simple syntax error - missing "do"
I don't understand ! I inserted the "do" before and it still returned an error!
Code:
while read i; do; if id $i > /dev/null 2>$1; then USER_LDAP_GROUPS="`id $i | cut -d\= -f4 | sed 's/([^()]*)//g;s/,/ /g'`"; echo $USER_LDAP_GROUPS; else echo "";fi; done < txt1
-bash: syntax error near unexpected token `;'

Jim yours works - thanks as always. I'm sure it's a syntax error on my part I'm just not readily seeing it. I tried several variations and still got an error.
# 6  
Old 07-11-2012
Quote:
Originally Posted by MaindotC
I don't understand ! I inserted the "do" before and it still returned an error!
Still returned what error?
# 7  
Old 07-11-2012
-bash: syntax error near unexpected token `;'
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