Getting part of variable

p0tw0r · 04-10-2005, 04:11 AM

I would like to echo some specific fields from a more complex variable.
In windows i used:

echo %var1%
result - my_var

echo %var1:~4,2%
result - ar

How do I do it in bash?

I posted it before on newbe and was suggested to use

cut -f

but I can't get it to work...

perfect_circle · 04-10-2005, 04:39 AM

Code:

skalkoto@darkstar:~/src$ var1=my_var
skalkoto@darkstar:~/src$ echo $var1
my_var
skalkoto@darkstar:~/src$ echo `expr substr $var1 5 2`
ar
skalkoto@darkstar:~/src$

for more info on string manipulation in bash read here:
http://www.tldp.org/LDP/abs/html/str...ipulation.html

perfect_circle · 04-10-2005, 04:52 AM

cut is for cutting the input into fields according to a delimiter ( -d) and displaying specific fields ( -f)

Code:

skalkoto@darkstar:~$ echo "my name is john" | cut -d " " -f 2,4
name john
skalkoto@darkstar:~$

p0tw0r · 04-11-2005, 02:25 AM

Thx! It works perfectly.