Script help please

Shell Programming and Scripting
1

Ok here is my code:

record_count_threshold=20
cd /usr/apps/data/output
ls -d *ROUTE* | while read line
do
echo "$line: `ls -ltr $line | grep -v DONE | wc -l`"
done > ${tmp_file}
echo
cat ${tmp_file} | awk '{if($2>'"${record_count_threshold}"'){print "Number of files backlogged: " $1 $2}}'

Ok so what I am trying to figure out is how to make this so that if there are no results that are over the threshold of 20 - that the output is:

NO BACKLOG

If there are results over 20 it displays those results

I was going to use a case statement but not sure how I am supposed to code that.

Thanks.

2 
ls -d *ROUTE*|while read -r dir;do if [ $(ls -1 $dir|grep -Ev 'DONE|total'|wc -l) -gt 20 ] ; then 
echo "$dir $(ls -ltr $dir|grep -Ev 'DONE|total')" ; else echo "$dir --> NO BACKLOG" ; fi ; done
3

Thank you. What is:

-gt

for?

And -E doesn't work for me on my syste.
Also what was the total for?

And last it works but it shows me the list of files where as I want just the count as there can be like 200 files in one directory.
thanks again!

---------- Post updated at 01:32 PM ---------- Previous update was at 01:10 PM ----------

i got it!
But i would still like to know the -gt and E

thank you!!!

4 
gt is greater than
-E use extended-regexp for grep

try change to

grep -Ev 'DONE|total'

to

egrep -v 'DONE|total'

regards
ygemici

5

Some basic questions.
What Operating System do you have?

uname -a

What Shell are you using?

echo "${SHELL}"

The script in Post #1 appears to be designed to look at all the directories which match the pattern *ROUTE* and count the number of files where the name of the file (not the contents) does not contain the string "DONE".

Hmm. The script as posted does not set a value for ${tmp_file}.

One idea without knowing what Shell you have:
Had a bit of fun avoiding the sort of syntax extensions we find in certain modern Shells.
Note the use of "ls -1" (one) rather than "ls -l" (ell).
I have assumed that there is only one depth of directory and removed "-r" from "ls". This may be wrong!

# Parameters
record_count_threshold=20
# Create temporary file including current process ID "$$" in the file name. 
tmp_file=/tmp/mytmpfile_$$
touch "${tmp_file}"    # Ensure file exists. This stops the count process failing if there is no backlog.
#
# Processing starts here
#
cd /usr/apps/data/output
#
# Look for directory names containing the string "ROUTE".
# Allow for no match by redirecting STDERR 
ls -1d *ROUTE* 2>/dev/null | while read dir
do
     # Make sure that this is a directory
     if [ -d "${dir}" ]
     then
            # Look for file matches
            # Allow for no match by redirecting STDERR
            ls -1 "${dir}" 2>dev/null | grep -v "DONE" | while read filename
            do
                   # Is the match a file?
                   if [ -f "${filename}" ]
                   then
                            # Output the filename so we can count them
                            echo "${filename}"
                   fi
            done
     fi
done > ${tmp_file}
#
# Count the number of matching files
backlog=`cat ${tmp_file}|wc -l`
# Compare with threshold.   -gt means Greater Than.
if [ ${backlog} -gt ${record_count_threshold} ]
then
        echo "Number of files backlogged: ${backlog}"
else
        echo "NO BACKLOG"
fi
#
# Clean up temporary file
rm "${tmp_file}"

Ps. Anybody who posts UUOC had better have a proven alternative piece of code.

6

UUOC!!!

backlog=$( wc -l ${tmp_file} | cut -d\  -f 1 )

:stuck_out_tongue:

7

@pludi & LOL
1) We don't know whether this Shell supports $( command ) syntax thought the redirect on the "done" line does suggests that it is ksh.
Anyway the "cat" solution is faster and easier to read.
2) Worth reading the "Useful Uses of Cat" web page.
Useful use of cat(1)
And Kernigan's document which is appended to that page. I really missed the "pip" command.

8

I know that cat has useful uses too (and probably would have written this line the same way you did), but if a challenge has been issued... :wink: