find content from a file - Shell Programming and Scripting

ust · #1 11-29-2007

root 0 0 5 0 11/09/07 08:18
root 0 0 6 0 11/09/07 08:56
root 0 0 7 0 11/22/07 11:09
user1 0 0 8 0 11/08/07 15:58
user2 0 0 9 0 11/30/07 08:37
user3 0 0 10 0 11/30/07 08:56
root 0 0 11 0 11/30/07 08:58
test
user4 0 0 12 0 11/30/07 09:02
user5 0 0 13 0 11/30/07 09:34
root 0 0 14 0 11/30/07 09:11
user6 0 0 15 0 11/30/07 09:26
user6 0 0 15 0 11/30/07 09 : 26
root 0 0 16 0 11/30/07 09:28

I have the above file that have many lines , I would like to find some content in the file and output to another file with the below condition as below format .

1. do not have the word root
2. only show column 1 , 3 & 6
3. add a line no. to each row
4. do not show the duplicate line ( as above , user6 is duplicate , then only show ONE time in the new output.

the output should be as below .

1 user1 0 11/08/07
2 user2 0 11/30/07
3 user3 0 11/30/07
4 user4 0 11/30/07
5 user5 0 11/30/07
6 user6 0 11/30/07

can advise what can i do ? thx

Smiling Dragon · #2 11-29-2007

Keeping in mind your comment about removing duplicates is ambiguous (do you mean only show one user6 per day or only one user6 - if so, which one?), here's a guess:

Code:

grep -v 'root' thefile | sort -u | awk '{ print NR,$1,$3,$6 }'

uvrakesh · #3 11-29-2007

Your requirement and the desired output is little different
only show column 1 , 3 & 6
then u must get 'test' in your output!!

This is the same line copied from smiling dragon, i have just introduced a check before printing.

grep -v 'root' sed.sh | sort -u | awk '{ if(user != $1){print NR,$1,$3,$6} ;user=$1 }'

RUV

summer_cherry · #4 11-29-2007

Hi,

This one should be ok.

Code:

sed '/root/d' filename | awk 'NF>=2' | sed 's/ : /:/' |sort -u | uniq -u | awk 'NF>=2 {print NR,$1,$3,$6}'

prowla · #5 11-29-2007

How about a method that uses more shell functionality:

Code:

lineno=0
ulast=blah
sort -u myfile | while read s
do
    set -- $s
    case $1 in
       $ulast) continue;;
       user*) ;;
       *) continue;;
    esac
    let lineno=lineno+1
    echo $lineno $1 $3 $6
    ulast=$1
done

venkatesh_sasi · #6 11-29-2007

Hi,

I have a very similar requirement. I'm storing the passwords in a flat file, when ever the user updates the password, I'm planning to write the new password along with some other values and delete the old one. I thought of doing it using "grep -v". Can anybody help me out how to pass a variable from C program to unix command. In my previous post, I got a reply from bakunin saying to use sprintf(). I tried the below code.
Code:
main()
{
char v_unix[10];
sprintf(v_unix,"%s",argv[1]);
system("cat filename | grep -v v_test > filename2");
}

This didn't work. Help me to accomplish this.

Thanks & Regards,
Venkatesh.

ust · #7 11-30-2007

Quote:

Originally Posted by summer_cherry

Hi,

This one should be ok.

Code:

sed '/root/d' filename | awk 'NF>=2' | sed 's/ : /:/' |sort -u | uniq -u | awk 'NF>=2 {print NR,$1,$3,$6}'

thx ,

this script is quite useful , the output as below,

1 user1 0 11/08/07
2 user2 0 11/30/07
3 user3 0 11/30/07
4 user4 0 11/30/07
5 user5 0 11/30/07
6 user6 0 11/30/07
7 user100 0 11/30/07
8 user1222 0 11/30/07

but if I want the output is more tidy as below , can advise what can i do ?

1 user1 0 11/08/07
2 user2 0 11/30/07
3 user3 0 11/30/07
4 user4 0 11/30/07
5 user5 0 11/30/07
6 user6 0 11/30/07
7 user100 0 11/30/07
8 user1222 0 11/30/07

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