Can we see output from ls -l too. That's a lower case L flag.
Regards,
Robin
find /usr/IBM/FileNet/BulkUploaderScript/$i/Log/SuccessLog \
\( -newer range_start -a \! -newer range_end \) \
| while read file
do
lines=`wc -l "$file"`---->i need number of lines in the file but i am getting filename also as shown below in the output
ls -l "$file" | read w x y z d1 d2 d3 rest--> it is not intializing variables but how to do that , i need basically lastmodification time rather than all infromation
echo "File $file has $lines lines and a date of $d1 $d2 $d3"
done
OUPUT:
File /usr/IBM/FileNet/BulkUploaderScript/LogsFolder/Log/SuccessLog/CSVSuccessLogHIPAACERTS7-6-2013 15-32-18.csv has 4
/usr/IBM/FileNet/BulkUploaderScript/LogsFolder/Log/SuccessLog/CSVSuccessLogHIPAACERTS7-6-2013 15-32-18.csv lines and a date of
Please give a sloution .
I am not able to see any output for that paticular ls-l
ls-l is not working , i mean it is not showing any output . please go through the above code in previous posts , if we do a minor change code will work . Please look into it and help
Do you have a space between ls & -l ?
When you say no output, do you mean absolutely nothing at all except that the prompt returns? You should get something, a total line, an error message, something surely.
Robin
find /usr/IBM/FileNet/BulkUploaderScript/$i/Log/SuccessLog \
\( -newer range_start -a \! -newer range_end \) \
| while read file
do
lines=`wc -l "$file"`-----
---->i need number of lines in the file but i am getting filename also as shown below in the output
`ls -l "$file" | read w x y z d1 d2 d3 rest`-----
it is not intializing variables but how to do that , i need basically lastmodification time ratherecho "File $file has $lines lines and a date of $d1 $d2 $d3"
echo $file>>$LOGFILE
done
OUPUT:
for all the lines above written
File /usr/IBM/FileNet/BulkUploaderScript/LogsFolder/Log/SuccessLog/CSVSuccessLogHIPAACERTS7-6-2013 15-32-18.csv has 4
/usr/IBM/FileNet/BulkUploaderScript/LogsFolder/Log/SuccessLog/CSVSuccessLogHIPAACERTS7-6-2013 15-32-18.csv lines and a date of
please suggest any alternative or proper solution
So, can we see the output of just ls -l on it's own in the correct directory. No further scripting, just the plain ls -l thanks.
Thanks,
Robin
find /usr/IBM/FileNet/BulkUploaderScript/$i/Log/SuccessLog \
\( -newer range_start -a \! -newer range_end \) \
| while read file
do
lines=`wc -l "$file"`
# details=`ls -l "$file" | read w x y z d1 d2 d3 rest`--->commented ,in this intailzing is problem
details=`ls -l "$file"`
echo $details
echo "File $file has $lines lines and a date of $d1 $d2 $d3"
echo $file>>$LOGFILE
done
OUTPUT:
-rwxrwxrwx 1 ce_admin ce_admin 718 Jun 7 15:32 /usr/IBM/FileNet/BulkUploaderScript/LogsFolder/Log/SuccessLog/CSVSuccessLogHIPAACERTS7-6-2013 15-32-18.csv
File /usr/IBM/FileNet/BulkUploaderScript/LogsFolder/Log/SuccessLog/CSVSuccessLogHIPAACERTS7-6-2013 15-32-18.csv has 4 /usr/IBM/FileNet/BulkUploaderScript/LogsFolder/Log/SuccessLog/CSVSuccessLogHIPAACERTS7-6-2013 15-32-18.csv lines and a date of
I need only last modified date of the file not access time .
And i am getting number of lines with filename but i need only number.
Any alternative or any modifications
:o Okay, I've spotted my first mistake. The output from ls -l "$file" has, of course, five colums to skip (permissions, links, owner, group & bytes) and what I suggested only has four dummy placeholders (w, x, y & z) :rolleyes:
Lets try adding dummy placeholder v to that:-
find /usr/IBM/FileNet/BulkUploaderScript/$i/Log/SuccessLog \
\( -newer range_start -a \! -newer range_end \) \
| while read file
do
lines=`wc -l "$file"`
details=`ls -l "$file" | read v w x y z d1 d2 d3 rest`
echo "File \"$file\" has \"$lines\" lines and a date of \"$d1\" \"$d2\" \"$d3\""
echo $file>>$LOGFILE
done
I've added in a lot of escaped quotes \" so that the output clearly defines each variable to see if I've missed something else.
Can you have a go and post the output in code tags - it makes it easier to read.
Thanks 
, and apologies :o
Robin
Still facing Issue that d1 ,d2,d3 not printing
Output::
Maybe we shoudl drop the details= bit:-
find /usr/IBM/FileNet/BulkUploaderScript/$i/Log/SuccessLog \
\( -newer range_start -a \! -newer range_end \) \
| while read file
do
lines=`wc -l "$file"`
ls -l "$file" | read v w x y z d1 d2 d3 rest
echo "File \"$file\" has \"$lines\" lines and a date of \"$d1\" \"$d2\" \"$d3\""
echo $file>>$LOGFILE
done
A schoolboy error of mine :o that makes these values would only be set within the sub-shell running the ls -l | read......
Can you actually copy & paste all the code block in, as you output didn't quite tally.
Does that help?
Robin
find /usr/IBM/FileNet/BulkUploaderScript/$i/Log/SuccessLog \
\( -newer range_start -a \! -newer range_end \) \
| while read file
do
lines=`wc -l "$file"`
ls -l "$file" | read v w x y z d1 d2 d3 rest
echo "File \"$file\" has \"$lines\" lines and a date of \"$d1\" \"$d2\" \"$d3\""
echo $file>>$LOGFILE
done
Variable "lines" reading filename also with the number of lines in the file .Majorly problem is d1,d2,d3 is not showing any output .What is the solution we have now .
Thanks
Raghavendra
Another variation to see where we're going:-
find /usr/IBM/FileNet/BulkUploaderScript/$i/Log/SuccessLog \
\( -newer range_start -a \! -newer range_end \) \
| while read file
do
lines=`grep -c "" "$file"`
ls -l "$file" | read v w x y z d1 d2 d3 rest
echo "File \"$file\" has \"$lines\" lines and a date of \"$d1\" \"$d2\" \"$d3\""
echo $file>>$LOGFILE
done
That should sort the counting. Testing here, the d1, d2 & d3 works fine for me:-
$ file=.profile
$ lines=`grep -c "" "$file"`
$ ls -l "$file" | read v w x y z d1 d2 d3 rest
$ echo "File \"$file\" has \"$lines\" lines and a date of \"$d1\" \"$d2\" \"$d3\""
File ".profile" has "12" lines and a date of "02" "May" "2012"
$
It would suggest that your code might have ls -1 | read .... instead of ls -l | read .... To clarify, you may have a number one where it needs a lower case L.
Can you have another go?
Robin
ksh reads the last part of the pipe in the current shell:
ls -l "$file" | read v w x y z d1 d2 d3 rest
sh,bash need a work-around:
read v w x y z d1 d2 d3 rest << _EOT
`ls -l "$file"`
_EOT
Or
set -- `ls -l "$file"`
d1=$6; d2=$7; d3=$8