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Set problem with spaces


 
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Top Forums Shell Programming and Scripting Set problem with spaces
# 8  
Old 09-07-2013
unfortunately, non of these are working.


here's how i'm doing it.

i have to set the variable this way:
Code:
set $(cat myvars)


then get the fields this way:

Code:
echo $* | awk/print whatever column i need

or
Code:
echo $@ | awk/print whatever column i need.

the spaces are being a major pain

ihave to do this on a linux/sunos host. linux hosts are ubuntu/centos/red hat.
# 9  
Old 09-07-2013
That is because in addition to your original problem specification you are using command substitution: $( ... ) which will cause the output of the command to be split into fields according to the IFS variable, which defaults to a space, a TAB and a newline. The same would go for variable expansion.

There is a difference between:
Code:
$ set -- hello 'cat dog walk' money elephant
$ echo "$2"
cat dog walk

and
Code:
$ var="hello 'cat dog walk' money elephant"
$ set -- $var
$ echo "$2"
'cat

in the first case there are 4 fields, and no field splitting is performed
in the second case there is field splitting after expansion of the variable "var", which results in 6 fields with the default IFS


--
You could try something like this:
Code:
$ oldIFS=$IFS
$ IFS="
"
$ set -- $(xargs -n1 < file)
$ IFS=$oldIFS
$ echo "$2"
cat dog walk
$


Last edited by Scrutinizer; 09-07-2013 at 01:11 PM..
This User Gave Thanks to Scrutinizer For This Post:
# 10  
Old 09-07-2013
Quote:
Originally Posted by Scrutinizer
That is because in addition to your original problem specification you are using command substitution: $( .. ) which will cause the output of the command to be split into fields according to the IFS variable, which defaults to a space, a TAB and a newline. The same would go for variable expansion.

There is a difference between:
Code:
$ set -- hello 'cat dog walk' money elephant
$ echo "$2"
cat dog walk

and
Code:
$ var="hello 'cat dog walk' money elephant"
$ set -- $var
$ echo "$2"
'cat

in the first case there are 4 fields, and no field splitting is performed
in the second case there is field splitting after expansion of the variable "var", which results in 6 fields with the default IFS


--
You could try something like this:
Code:
$ oldIFS=$IFS
$ IFS="
"
$ set -- $(xargs -n1 < infile2443)
$ IFS=$oldIFS
$ echo "$2"
cat dog walk
$

this worked. thank you so much!!!!
# 11  
Old 09-07-2013
This should work as well:
Code:
IFS="
" set -- $(xargs -n1 < infile2443)
echo "$2"

# 12  
Old 09-07-2013
That will not work, set does not use IFS.
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