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Shell Programming and Scripting

Don't display file name

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Top Forums Shell Programming and Scripting Don't display file name
# 1  
Old 04-17-2012
Bug Don't display file name
Hi

I have a directory where i have 10 -12 files, i want to store word count in seperate file without any file name.

i have written the logic which needs to be corrected
Code:
 filename.txt
for i in /backup/temp/rajesh/12Apr2012_South *.LOG
do
#echo "File size of $i is"
wc -l $i
done>lines.txt

Thanks

Rajesh
Last edited by methyl; 04-17-2012 at 10:57 AM.. Reason: please use code tags
# 2  
Old 04-17-2012
By redirecting the file to wc, rather than having wc open the file, wc will not print a filename.

Code:
for file in /backup/temp/*.LOG; do
    wc -l < "$file"
done > "lines.txt"

# 3  
Old 04-17-2012
Quote:
wc -l $i
Try
Code:
wc -l < $i

# 4  
Old 04-17-2012
Hi,

try this:

Code:
wc -l  /anywhere/file*.log |cut -d " " -f 1 > result.txt

Moderator's Comments:
Mod Comment Use code tags instead of quote tags, thank you
Last edited by Scrutinizer; 04-17-2012 at 01:00 PM..
# 5  
Old 04-17-2012
GNU sed if there are not too many files:
Code:
sed -sn '$=' file*

--
But the OP said he wanted to count words (wc -w) and we are all counting lines..

--
@pokerino: cut will not do here, because it uses single characters as separators and also the last number will be the total...
# 6  
Old 04-17-2012
correct:
Code:
wc -w *|awk -F " " '{print $1 }' > result.txt

Last edited by Scrutinizer; 04-17-2012 at 04:25 PM..
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