Cut Help

metallica1973 · 12-05-2007, 12:49 PM

How can use the cut command to cut the last two octets or hostname from a string like for example:

PHP Code:



S00CBBF.aa.bb.ccc.edu

To

PHP Code:



S00CBBF.aa.bb

Thanks

schneidz · 12-05-2007, 01:12 PM

i dunno' if there is a simpler way but this works:

Code:

echo S00CBBF.aa.bb.ccc.edu | rev | cut -d . -f 3- | rev

pixellany · 12-05-2007, 01:12 PM

Why not SED. This is crude, but works:

sed 's/\....\....$//' filename

Translation: find a literal ".", any 3 characters, another literal ".", 3 more characters, and then the end of the line. Replace by an empty string.

metallica1973 · 12-05-2007, 01:13 PM

I am using Sun Solaris 10. Is that command 'rev' still valid?

schneidz · 12-05-2007, 01:32 PM

Quote:

Originally Posted by metallica1973

I am using Sun Solaris 10. Is that command 'rev' still valid?

i guess not:

Code:

Sun Microsystems Inc.   SunOS 5.8       Generic Patch   February 2004
schneidz@lq:/temp> echo hello-world | rev
ksh: rev:  not found

try compiling yourself one of these:

Code:

schneidz@lq:/temp> cat ssh-rev.c
#include <string.h>
#include <stdio.h>


char *str_rev(char *s)
    {
    char *p=s;
    char *q =s;
    char swap;
    if (*s)
    {
        q=strchr(q,'\0');
        while (--q > p)
        {
            swap = *q;
            *q = *p;
            *p = swap;
            ++p;
        }
    }
    return s;
}

main(int argc, char * argv[])
{
 char * loc1;
 char * loc2;
 int locdiff, argv2strlen;
 int pos = 0;

  argv[1] = str_rev(argv[1]);
  loc1 = strchr(argv[1], argv[1][0]);
  loc2 = strchr(argv[1], '/');
  locdiff = loc2 - loc1;
  argv[1][locdiff] = '\0';
  printf(" %s \n", str_rev(argv[1]));
}

ilikejam · 12-05-2007, 02:20 PM

awk for the win.

Code:

awk -F'.' '{for (i=1;i<=NF-2;i++) {if (i<NF-2) printf $i"."; else print $i}}'

Dave

chrism01 · 12-05-2007, 04:50 PM

echo S00CBBF.aa.bb.ccc.edu |cut -d'.' -f1-3

gives
S00CBBF.aa.bb

too easy

and shorter / clearer than awk

ilikejam · 12-05-2007, 04:53 PM

Quote:

Originally Posted by chrism01

echo S00CBBF.aa.bb.ccc.edu |cut -d'.' -f1-3

gives
S00CBBF.aa.bb

too easy

and shorter / clearer than awk

Bzzzzt. The OP wanted rid of the last two fields. What if you don't know in advance how many fields there are in the string?

unSpawn · 12-05-2007, 05:21 PM

Quote:

Originally Posted by ilikejam

Bzzzzt. The OP wanted rid of the last two fields. What if you don't know in advance how many fields there are in the string?

OK, BASH prolly isn't what you're looking for but w/o 'cut': i="S00CBBF.aa.bb.ccc.edu"; i=(${i//./ }); echo "${i[0]}.${i[1]}"

ilikejam · 12-05-2007, 05:26 PM

No joy, I'm afraid:

Code:

[0 dave@cronus ~]$ i="S00CBBF.aa.bb.ccc.edu"; i=(${i//./ }); echo "${i[0]}.${i[1]}"
S00CBBF.aa

Should give: S00CBBF.aa.bb , and you'd still have to know in advance how many fields there are in any case.

awk's still in front. Go awk! etc.

Anyone got any perl?

Dave

ghostdog74 · 12-05-2007, 05:28 PM

Code:

# echo S00CBBF.aa.bb.ccc.edu | nawk 'gsub(/\.[a-zA-Z0-9]+\.[a-zA-Z0-9]+$/,"")'

unSpawn · 12-05-2007, 05:38 PM

heh. way too much work: i="S00CBBF.aa.bb.ccc.edu"; i=(${i//./ }); n=${#i[@]}; unset i[$[$n-1]] i[$[$n-2]]; i=${i[*]}; echo ${i// /.}

chrism01 · 12-05-2007, 06:26 PM

Well, given the OP's example input, 'removing the last 2 fields' and 'keeping the first 3' both give his example answer. The qn is, what does he REALLY need??

metallica1973 · 12-05-2007, 07:04 PM

Everyone please read my original post so that all of you can understand my situation.

http://www.linuxquestions.org/questi...-shell-604725/

The output will always be in this format:

PHP Code:



S00F3456.aa.bb.cc.dd.ee.fff.edu

or

PHP Code:



S00F3456.aa.bb.cc.dd.fff.edu

We use unix for DNS and we have to use the fully qualified name so what I want is from the output above, logic that can distinguish different fields in the output and remove the last 2 octects, always. so I would need an output like

PHP Code:



S00F3456.bb.cc.dd

or

PHP Code:



S00F3456.bb.cc.dd.ee

the last will need to truncated no matter how long the output is!

ilikejam · 12-05-2007, 07:20 PM

So, basically, you need a function which will remove the last two 'between-dots' fields from your hostname string.

In that case,

Code:

awk -F'.' '{for (i=1;i<=NF-2;i++) {if (i<NF-2) printf $i"."; else print $i}}'

will do it, e.g.

Code:

[0 dave@cronus ~]$ echo S00F3456.aa.bb.ccc.edu | awk -F'.' '{for (i=1;i<=NF-2;i++) {if (i<NF-2) printf $i"."; else print $i}}'
S00F3456.aa.bb
[0 dave@cronus ~]$ echo S00F3456.bb.ccc.edu | awk -F'.' '{for (i=1;i<=NF-2;i++) {if (i<NF-2) printf $i"."; else print $i}}'
S00F3456.bb
[0 dave@cronus ~]$ echo S00F3456.aa.bb.xyz.abc.ccc.edu | awk -F'.' '{for (i=1;i<=NF-2;i++) {if (i<NF-2) printf $i"."; else print $i}}'
S00F3456.aa.bb.xyz.abc

Dave