split string on file delimeter

Jeroen1000 · 10-05-2009, 01:45 AM

I gave it a shot but something still isn't entirely correct

Code:

# Next split each record of data to obtain the ip address.
old_IFS=$IFS
IFS=$':'
ip="a:b:c"
for s in $ip; 

do echo $s; done

I only want the letter "a" in variable s (no spaces and no carriage return either). But when I echo s it displays:

a
b
c

A little help will surely get me there

.

kbp · 10-05-2009, 02:01 AM

You can use cut, then you don't have to muck about with the IFS
eg.

Code:

# Next split each record of data to obtain the ip address.
ip="a:b:c"
for s in $ip; 
do
echo $s | cut -d: -f1
done

cheers

lutusp · 10-05-2009, 02:02 AM

Quote:

Originally Posted by Jeroen1000

I gave it a shot but something still isn't entirely correct

Code:

# Next split each record of data to obtain the ip address.
old_IFS=$IFS
IFS=$':'
ip="a:b:c"
for s in $ip; 

do echo $s; done

I only want the letter "a" in variable s (no spaces and no carriage return either). But when I echo s it displays:

a
b
c

A little help will surely get me there

.

1. Don't reset IFS. That's not necessary in modern times.

Code:

$ ip="a:b:c"
$ echo ${ip%%:*}
a

But it would help if you would say what you want, not what you don't want.

Jeroen1000 · 10-05-2009, 02:25 AM

both methods work perfectly guys

Lutusp, incidentally, I was just mocking about with the method you use.

Bash is really chinese to me compared to any other language

.

Cheers,

Jeroen

Jeroen1000 · 10-05-2009, 03:05 AM

Dumb ass follow up question

When I use lutusp's method I can do

Code:

var=${ip%%:*}
echo $var

But because I want to learn bash, I also decided to try the other method:

The code below is not allowed

Code:

var=echo $s | cut -d: -f1
echo $var

I've tried various things, like adding () around the echo statement etc. Can someone instruct me why this assignment is not allowed?

Jeroen1000 · 10-05-2009, 03:28 AM

I found the answer, don't know why it has to be this way but this is the solution:

Code:

user=`echo ${data[0]} | cut -d: -f2`

catkin · 10-05-2009, 03:30 AM

Quote:

Originally Posted by Jeroen1000

Code:

var=echo $s | cut -d: -f1
echo $var

I've tried various things, like adding () around the echo statement etc. Can someone instruct me why this assignment is not allowed?

The line tells the shell "assign echo to variable var for the duration of running the following command which is ... run the command which is the value of $s and pipe its stdout to the stdin of the cut command.

What you want to do is assign the output from "echo $s | cut -d: -f1" to variable var. This will do it

Code:

var="$(echo $s | cut -d: -f1 )"

The $( ) makes the stdout from the contained line available as a string. The " " around it makes it a single string and prevents it being split at the first whitespace character and so ensures the whole of the command output is assigned to $var.

gnashley · 10-05-2009, 08:35 AM

To avoid the carriage return use 'echo -n ....'