probably a very basic 'sed' question

[MHThomas]

Hi,
I wonder if any body can help me with a simple data manipulation problem. I am using Korn shell on AIX and need to select the first section of a variable name, keeping everything before the first '.' and dumping everything afterwards. As an example I would want the following script (with the blank filled in):

#!/bin/ksh
x="aaa.bbb.ccc.com"
y=#######unknown (sed?) statement#######
echo output: $y


To produce the following results:

output: aaa

I'm sure this is simple, but I just can't get any joy.

Cheers,

Mark T.

 

[Ygor]

y=$(print $x|cut -d. -f1)

 

[Salem]

One way is

y=`echo $x | awk -F. '{print $1}'`

 

[MHThomas]

Thank you both for your amazingly quick responses. I'll buy you both a pint the next time I see you

Cheers,

Mark T.

 

[vgersh99]

#!/bin/ksh

x="aaa.bbb.ccc.com"
y="${x%%.*}"

echo "[$x] [$y]"


vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[pavNell]

If you wanted to use sed...

y=`echo $x | sed 's/^\(.*\)\.*$/\1/'`

 

[vgersh99]

sed's regex are greedy.

y=`echo $x | sed 's/^\([^.][^.]*\).*$/\1/'

vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+