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variables in ksh

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# 1  
Old 04-13-2002
variables in ksh

I'm new to unix scripting. How would I go about pulling the first 3 characters from a variable in ksh and storing in another variable? Thanks.
# 2  
Old 04-13-2002
Several ways, of course. One popular way is:
var2=`echo $var1 | cut -c1-3`
and some will pipe it through sed, but I prefer to use expr because it is just one command, no pipe:
var2=`expr "$var1" : "\(...\)"`

Whatever command to use, we are "nesting" the command string in back-quotes. The back-quoted command will be replaced in the command line by its output.

In ksh, I think the most efficient way (do man ksh), when you can use it:

var2=${var1#pattern} (remove one occurent from front)
var2=${var1##pattern} (remove mult occurrences from front)
var2=${var1%pattern} (remove one occurrence from end)
var2=${var1%%pattern} (remove mult occurrences from end)

But in this case, we want to remove, from the end, all but the first 3 characters, and I could not come up with an expression to represent that.
# 3  
Old 04-17-2002
Try this:
$ var=123456
$ var=${var##???}
$ echo $var

Hope this helps!
# 4  
Old 04-17-2002
Yep, that will delete the first 3 characters, but steve6368 was wanting to pull (retain) the first 3 characters into another variable.
# 5  
Old 04-17-2002
Oops, in that case it should be :
$ var=123456
$ var=${var%%???}
$ echo $var

# 6  
Old 04-17-2002

What are the functionality of ## and %% ? Any more special symbols we can use ?

# 7  
Old 04-17-2002
LivinFree, that will delete last 3 characters, and would retain exactly 3 only when your original variable is exactly 6 character. If your original variable is 10 characters, that solution pulls the first 7.

I could not think of a way to retain just the first 3 using that construct, as I mentioned in my first reply. The %% expression would need to represent "all but the first 3 characters".

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