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Open variable conatin name

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# 1  
Old 02-08-2013
Open variable conatin name
hi All,

I have a variable c which collects logs path name like "xyz.date.log".
Now with that Path name i want to open (cat) the file in log directory and then grep for word like bytes then awk for $6 .
Code:
cd /log/

Code:
D=`cat $C |grep -i wrote |awk 'print $6;}'


does it makes any sense ..Smilie btw its not gv any output.
# 2  
Old 02-08-2013
Quote:
Originally Posted by netdbaind;302768083[CODE
D=`cat $C |grep -i wrote |awk 'print $6;}'[/CODE]
that's useless use of cat and you have done many mistakes above.
try sth like this..

Code:
D=$(grep -i wrote $C | awk '{print $6}')

# 3  
Old 02-08-2013
Thanks for Reply Pamuand correction also .

no o/p gettng with this code.

c has only the file name xyz.date.log we need to more or cat /log/xyz.date.log and then grep and awk.

---------- Post updated at 02:50 AM ---------- Previous update was at 02:48 AM ----------

sorry for typo.

*Thanks for Reply PAMU and for correction also.
# 4  
Old 02-08-2013
You could do this in one sweep with awk:
Code:
$ D=$(awk '/[wW][rR][oO][tT][eE]/ {print $6}'  "$c")
or
$ D=$(awk 'toupper($0) ~ /WROTE/ {print $6}'  "$c")

# 5  
Old 02-08-2013
Code:
D=$(awk 'tolower($0) ~ /wrote/ {print $6}'  "$C")

Edit: See that RudiC posted nearly the same.
# 6  
Old 02-08-2013
Rudic,

code hanged it seems...no o/p Smilie
# 7  
Old 02-08-2013
- does your file contain the string "wrote" in any combination of upper/lower case?
- does $c translate to that file? (I see the other contributors have $C - you used a lower case c in your post#1.) Try replacing $c with $C.
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