Bash script for printing folder names and their sizes
1. The problem statement, all variables and given/known data:
The task is to create a script that would reproduce the output of 'du' command, but in a different way: what 'du' does is:
and what is needed is
We need to show only 10 folders which are the biggest ones.
2. Relevant commands, code, scripts, algorithms:
Nothing special was stated here, so any solution that can still be called a bash script is good enough to be used.
3. The attempts at a solution (include all code and scripts):
I've tried several variants and finally came up with this:
It does the job, but works too slowly to be used on practice: for my '/usr' folder it takes 7 minutes and 30 seconds to complete, while the 'du' with filters ('du /usr | sort -nr -k1 | head -10') does everything for 2 minutes and 30 seconds.
4. Complete Name of School (University), City (State), Country, Name of Professor, and Course Number (Link to Course):
Linux System Administration course, ITExpert educational centre, Minsk, Belarus.
Hi ,
I'm trying to list the files and output is written to a file. But when I execute the command , the output file is being listed. How to exclude it ?
/tmp
file1.txt
file2.txt
ls -ltr |grep -v '-' | awk print {$9, $5} > output.txt
cat output.txt
file1.txt
file2.txt
output.txt (8 Replies)
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Good day, everyone!
I'm very new to bash scripting. Our teacher gave us a task to create a script that basically does the same job the 'du' command does, with the difference that 'du' command gives an output in the form of
<size> <folder name>and what we need is
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