Question

I am a beginner in shell. Please help me with this question.

question:

In this assignment, you will research into a few standard Unix utilities and leverage them in a shell script to perform a simple but useful task that's right up the alley of Unix shell scripting.

As usual, you should aim for reasonably efficient algorithms and reasonably organized, comprehensible code.

Consolidating log files

My Internet chat software recording my conversations into daily log files, each named by date, e.g., “2020-06-14.log”. Naturally, as months go by, this becomes unwieldily too many files.

If I can merge daily log files into monthly log files, that will be great. But of course, I want to preserve the dates in another way. Help me by writing a shell script for this, call it “stampmerge.sh”. Here is the precise specification:

For each filename of the form “yyyy-mm-dd.log” in the current directory, in alphabetical order (if you use a shell pattern, it's already ordered for you):

Append the content to “yyyy-mm.log”, but with this addition: Each line is prefixed with “yyyy-mm-dd” followed by a space. Example: If 2020-06-14.log has these lines:

Hello everyone in CSCB09
Bye!

then append these lines to 2020-06.log:

2020-06-14 Hello everyone in CSCB09
2020-06-14 Bye!

Moreover! Putting a space between the date and the original line is only the default. If I provide a command line argument, use that instead; you may assume that it's just one character. Example: if I run “sh stampmerge.sh :”, then the lines to be appended are:

2020-06-14:Hello everyone in CSCB09
2020-06-14:Bye!

Sample input (including a pre-existing monthly file that you must append to, not overwrite) and sample expected output 2020-06.log.expected (using the default space) are provided.

I will test your script with sh; bash extra features will not be supported. I may test with no permission to create new files—you don't need it.

Useful utilities

cut can help extract “2020-06” and “2020-06-14” from “2020-06-14.log”.

yes, head, paste can help with prefixing a string to every line. But you also need to know how many lines are in the daily file, so use wc.

Answer 1

Hi,

Please find the code below.

echo $PWD
FILES=/home/*.log
for f in $FILES
do
echo "Processing $f file..."
# take action on each file. $f store current file name
cat $f
  
abc=$(echo $f | cut -d "." -f1 |rev|cut -d"-" -f2-|rev|cut -d'/' -f3-)
echo # from this i have got just year and month part of the log file name.
echo $abc
  
# this wil attach each line of the file with year and month.
if [ $# -eq 0 ]
then
echo "No arguments supplied"
while IFS= read -r line; do
echo "${abc} $line"
done < "$f" > $abc.log
else
echo "arguments supplied"
while IFS= read -r line; do
echo "${abc}${1}$line"
done < "$f" > $abc.log
fi

done
  
  

code snapshot:

4 11 13 main.bash 2020-06-19.log 2020-06.log! 1 echo $PWD 2 FILES=/home/*.log 3 for f in $FILES do 5 echo "Processing $f file..." 6 # take action on each file. $f store current file name 7 cat $f 8 9 abc=$(echo $f cut -d "." -f1 revcut -d"-" --2- |revcut -d/' -f3-) 10 echo # from this i have got just year and month part of the log file name. echo $abc 12 # this wil attach each line of the file with year and month. 14 if [ $# -eq 0 ] 15 then 16 echo "No arguments supplied" 17 while IFS= read -r line; do 18 echo "${abc} $line" 19 done < "$f" > $abc.log 20 else 21 echo "arguments supplied" 22 while IFS= read -r line; do 23 echo "${abc}${1}$line" 24 done < "$f" > $abc.log| 25 fi 26 27 done

output :

without command line argument:

main.bash 2020-06-19.log 2020-06.log 2020-06 Hello everyone in CSCBO9 2 2020-06 Bye! 3 1

output attached.with command line argument: ":"

main.bash 2020-06-19.log 2020-06.log! 1 2020-06:Hello everyone in CSCB09 2 2020-06:Bye! 3

Thanks,

kindly upvote