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# 8  
Old 01-06-2011
You need one more backslash with the old command substitution command:

Code:
export a=`awk -F'\\\^A' '{ print $1 }' file1.sh`

But, of course, you should use the new one $( ... ), if your shell supports it.
These 2 Users Gave Thanks to radoulov For This Post:
gc_sw michaelrozar17
# 9  
Old 01-06-2011
Quote:
Originally Posted by radoulov
You need one more backslash with the old command substitution command:

Code:
export a=`awk -F'\\\^A' '{ print $1 }' file1.sh`

But, of course, you should use the new one $( ... ), if your shell supports it.
Thanks radoulov. I have two question wrt the field separator. If the field separator is ^A in the file, why do we have 2 backslashes, i thought one would be enough.If 2 backslashes it mean literally FS= \^A

Wrt your recent post..How does adding another one backslash to the existing 2 backslashes work in the script.?
This User Gave Thanks to michaelrozar17 For This Post:
gc_sw
# 10  
Old 01-06-2011
Quote:
I have two question wrt the field separator.
If the field separator is ^A in the file, why do we have 2 backslashes, i thought one would be enough.
If 2 backslashes it mean literally FS= \^A
Good questions.
From the GNU awk manual (p 46,-47 from rhe pdf version)

Quote:
... with Unix awk and gawk, the
assignment ‘FS = "\.."' assigns the character string ".." to FS (the backslash is stripped).
This creates a regexp meaning “fields are separated by occurrences of any two characters.”
If instead you want fields to be separated by a literal period followed by any single character,
use ‘FS = "\\.."'.
So in the assignment you loose one backslash and in order to get \^ you need \\^.

Quote:
How does adding another one backslash to the existing 2 backslashes work in the script.?
This is special to the old-style backquote form of substitution `...`, it just removes a \
when it's followed by another \, from the manual pages of bash:

Quote:
When the old-style backquote form of substitution is used, backslash retains its literal meaning except when
followed by $, `, or \. The first backquote not preceded by a backslash terminates the command substitution.
When using the $(command) form, all characters between the parentheses make up the command; none are treated
specially.
Code:
% printf '%s\n'  `printf '%s' '\\a'`  
\a
% printf '%s\n'  $(printf '%s' '\\a') 
\\a

# 11  
Old 01-06-2011
Quote:
Originally Posted by radoulov
So in the assignment you loose one backslash and in order to get \^ you need \\^.
Here the delimiter of file file1.sh is ^A..am i correct? If so then why we have the delimiter literally \^A as you have said above.
# 12  
Old 01-06-2011
Quote:
Originally Posted by michaelrozar17
Here the delimiter of file file1.sh is ^A..am i correct? If so then why we have the delimiter literally \^A as you have said above.
The awk FS contains either a single character or a multi-character regular expression.
Therefore the ^ is special (a zero-width match at the beginning of the string) and
if you need to match a literal ^, you need to escape it.
Last edited by radoulov; 01-06-2011 at 07:33 AM..
# 13  
Old 01-06-2011
Is the field separator a literal "^A" or SOH (ASCII Start of Header)? If it is SOH, I do not see any reason for the guarding backslashes.

Both
Code:
$ cut -f2 -d'^A' infile
34

and
Code:
$ a=$(awk -F'^A' '{ print $2 }' infile); print $a
34

work perfectly well for me if SOH is entered as CTRL-V,CTRL-A.
# 14  
Old 01-06-2011
Quote:
Originally Posted by fpmurphy
Is the field separator a literal "^A" or SOH (ASCII Start of Header)? If it is SOH, I do not see any reason for the guarding backslashes.

Both
Code:
$ cut -f2 -d'^A' infile
34

and
Code:
$ a=$(awk -F'^A' '{ print $2 }' infile); print $a
34

work perfectly well for me if SOH is entered as CTRL-V,CTRL-A.
It's not the case, please read post #4.
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