How to call a function while using find in bash? - bash

So my objective here is to print a small graph, followed by the file size and the path for the 15 largest files. However, I'm running into issues trying to call the create_graph function on each line. Here's what isn't working
find $path -type f | sort -nr | head -$n | while read line; do
size=$(stat -c '%s' $line)
create_graph $largest $size 50
echo "$size $line"
done
My problem is that it isn't sorting the files, and the files aren't the n largest files. So it appears my "while read line" is messing it all up.
Any suggestions?

The first command,
find $path -type f
just prints out file names. So it can't sort them by size. If you want to sort them by size, you need to make it print out the size. Try this:
find $path -type f -exec du -b {} \; | sort -nr | cut -f 2 | head -$n | ...
Update:
Actually, only the first part of that seems to do everything you want from it:
find $path -type f -exec du -b {} \; | sort -nr | head -$n
will print out a table with size and filename, sorted by file size, and limited to $n rows.
Of course I don't know what the create_graph does.
Explanation:
find $path -type f -exec du -b {} \;
Find all files (not directories or links) in ${path} or its subdirectories, and execute the command du -b <file> on each.
du -b <file>
will output the size of the file (disk usage). See man du for details.
This will produce something like this:
8880 ./line_too_long/line.o
4470 ./line_too_long/line.f
934 ./random/rand.f
9080 ./random/rand
23602 ./random/monte
7774 ./random/monte.f90
13610 ./format/form
288 ./format/form.f90
411 ./delme.f90
872 ./delme_mod.mod
9029 ./delme
So for each file, it prints the size (-b for 'in bytes').
Then you can do a numerical sort on that.
$ find . -type f -exec du -b {} \; | sort -nr
23602 ./random/monte
13610 ./format/form
9080 ./random/rand
9029 ./delme
8880 ./line_too_long/line.o
7774 ./random/monte.f90
4470 ./line_too_long/line.f
934 ./random/rand.f
872 ./delme_mod.mod
411 ./delme.f90
288 ./format/form.f90
And if you then cut it off after the first, say five entries:
$ find . -type f -exec du -b {} \; | sort -nr | head -5
23602 ./random/monte
13610 ./format/form
9080 ./random/rand
9029 ./delme
8880 ./line_too_long/line.o
Some idea to put that back together:
find . -type f -exec du -b {} \; | sort -nr | head -$n | while read line; do
size=$(cut -d ' ' -f 1 <<< $line)
file=$(cut -d ' ' -f 2 <<< $line)
create_graph $largest $size 50
echo $line
done
Note that I have no idea what create_graph is or what $largest contains. I took that straight out of your script.

Related

How to count files in subdir and filter output in bash

Hi hoping someone can help, I have some directories on disk and I want to count the number of files in them (as well as dir size if possible) and then strip info from the output. So far I have this
find . -type d -name "*,d" -print0 | xargs -0 -I {} sh -c 'echo -e $(find "{}" | wc -l) "{}"' | sort -n
This gets me all the dir's that match my pattern as well as the number of files - great!
This gives me something like
2 ./bob/sourceimages/psd/dzv_body.psd,d
2 ./bob/sourceimages/psd/dzv_body_nrm.psd,d
2 ./bob/sourceimages/psd/dzv_body_prm.psd,d
2 ./bob/sourceimages/psd/dzv_eyeball.psd,d
2 ./bob/sourceimages/psd/t_zbody.psd,d
2 ./bob/sourceimages/psd/t_gear.psd,d
2 ./bob/sourceimages/psd/t_pupil.psd,d
2 ./bob/sourceimages/z_vehicles_diff.tga,d
2 ./bob/sourceimages/zvehiclesa_diff.tga,d
5 ./bob/sourceimages/zvehicleswheel_diff.jpg,d
From that I would like to filter based on max number of files so > 4 for example, I would like to capture filetype as a variable for each remaining result e.g ./bob/sourceimages/zvehicleswheel_diff.jpg,d
I guess I could use awk for this?
Then finally I would like like to remove all the results from disk, with find I normally just do something like -exec rm -rf {} \; but I'm not clear how it would work here
Thanks a lot
EDITED
While this is clearly not the answer, these commands get me the info I want in the form I want it. I just need a way to put it all together and not search multiple times as that's total rubbish
filetype=$(find . -type d -name "*,d" -print0 | awk 'BEGIN { FS = "." }; {
print $3 }' | cut -d',' -f1)
filesize=$(find . -type d -name "*,d" -print0 | xargs -0 -I {} sh -c 'du -h
{};' | awk '{ print $1 }')
filenumbers=$(find . -type d -name "*,d" -print0 | xargs -0 -I {} sh -c
'echo -e $(find "{}" | wc -l);')
files_count=`ls -keys | nl`
For instance:
ls | nl
nl printed numbers of lines

How can I count the number of words in a directory recursively?

I'm trying to calculate the number of words written in a project. There are a few levels of folders and lots of text files within them.
Can anyone help me find out a quick way to do this?
bash or vim would be good!
Thanks
use find the scan the dir tree and wc will do the rest
$ find path -type f | xargs wc -w | tail -1
last line gives the totals.
tldr;
$ find . -type f -exec wc -w {} + | awk '/total/{print $1}' | paste -sd+ | bc
Explanation:
The find . -type f -exec wc -w {} + will run wc -w on all the files (recursively) contained by . (the current working directory). find will execute wc as few times as possible but as many times as is necessary to comply with ARG_MAX --- the system command length limit. When the quantity of files (and/or their constituent lengths) exceeds ARG_MAX, then find invokes wc -w more than once, giving multiple total lines:
$ find . -type f -exec wc -w {} + | awk '/total/{print $0}'
8264577 total
654892 total
1109527 total
149522 total
174922 total
181897 total
1229726 total
2305504 total
1196390 total
5509702 total
9886665 total
Isolate these partial sums by printing only the first whitespace-delimited field of each total line:
$ find . -type f -exec wc -w {} + | awk '/total/{print $1}'
8264577
654892
1109527
149522
174922
181897
1229726
2305504
1196390
5509702
9886665
paste the partial sums with a + delimiter to give an infix summation:
$ find . -type f -exec wc -w {} + | awk '/total/{print $1}' | paste -sd+
8264577+654892+1109527+149522+174922+181897+1229726+2305504+1196390+5509702+9886665
Evaluate the infix summation using bc, which supports both infix expressions and arbitrary precision:
$ find . -type f -exec wc -w {} + | awk '/total/{print $1}' | paste -sd+ | bc
30663324
References:
https://www.cyberciti.biz/faq/argument-list-too-long-error-solution/
https://www.in-ulm.de/~mascheck/various/argmax/
https://linux.die.net/man/1/find
https://linux.die.net/man/1/wc
https://linux.die.net/man/1/awk
https://linux.die.net/man/1/paste
https://linux.die.net/man/1/bc
You could find and print all the content and pipe to wc:
find path -type f -exec cat {} \; -exec echo \; | wc -w
Note: the -exec echo \; is needed in case a file doesn't end with a newline character, in which case the last word of one file and the first word of the next will not be separated.
Or you could find and wc and use awk to aggregate the counts:
find . -type f -exec wc -w {} \; | awk '{ sum += $1 } END { print sum }'
If there's one thing I've learned from all the bash questions on SO, it's that a filename with a space will mess you up. This script will work even if you have whitespace in the file names.
#!/usr/bin/env bash
shopt -s globstar
count=0
for f in **/*.txt
do
words=$(wc -w "$f" | awk '{print $1}')
count=$(($count + $words))
done
echo $count
Assuming you don't need to recursively count the words and that you want to include all the files in the current directory , you can use a simple approach such as:
wc -l *
10 000292_0
500 000297_0
510 total
If you want to count the words for only a specific extension in the current directory , you could try :
cat *.txt | wc -l

Getting file size in bytes with bash (Ubuntu)

Hi, i'm looking for a way to output a filesize in bytes. Whatever i try i will get either 96 or 96k instead of 96000.
if [[ -d $1 ]]; then
largestN=$(find $1 -depth -type f | tr '\n' '\0' | du -s --files0-from=- | sort | tail -n 1 | awk '{print $2}')
largestS=$(find $1 -depth -type f | tr '\n' '\0' | du -h --files0-from=- | sort | tail -n 1 | awk '{print $1}')
echo "The largest file is $largestN which is $largestS bytes."
else
echo "$1 is not a directory..."
fi
This prints "The largest file [file] is 96k bytes"
there is -b option for this
$ du -b ...
Looks like you're trying to find the largest file in a given directory. It's more efficient (and shorter) to let find do the heavy lifting for you:
find $1 -type f -printf '%s %p\n' | sort -n | tail -n1
Here, %s expands to the size in bytes of the file, and %p expands to the name of the file.

Print an ordered list of files based on files size in bash

I made the following script to find files based on a 'find' command and then print out the results:
#!/bin/bash
loc_to_look='./'
file_list=$(find $loc_to_look -type f -name "*.txt" -size +5M)
total_size=`du -ch $file_list | tail -1 | cut -f 1`
echo 'total size of all files is: '$total_size
for file in $file_list; do
size_of_file=`du -h $file | cut -f 1`
echo $file" "$size_of_file
done
...which give me output like:
>>> ./file_01.txt 12.0M
>>> ./file_04.txt 24.0M
>>> ./file_06.txt 6.0M
>>> ./file_02.txt 6.2M
>>> ./file_07.txt 84.0M
>>> ./file_09.txt 55.0M
>>> ./file_10.txt 96.0M
What I would like to do first, though, is sort the list by file size before printing it out. What is the best way to go about doing this?
Easy to do if you grab the file size in bytes, just pipe to sort
find $loc_to_look -type f -name "*.txt" -size +5M -printf "%f %s\n" | sort -n -k 2
If you wanted to make the file sizes print in MB, you could finally pipe to awk:
find $loc_to_look -type f -printf "%f %s\n" | sort -n -k 2 | awk '{ printf "%s %.1fM\n", $1, $2/1024/1024}'

How sort find result by file sizes

How can I sort by file size the results of the find command?
I trying to sort the result of this find command:
find ./src -type f -print0
I don't want the size of directories, I need the files relative paths sorted by size only.
Here is how to do using find command:
find . -type f -exec ls -al {} \; | sort -k 5 -n | sed 's/ \+/\t/g' | cut -f 9
Here is how to do using recursive ls command:
ls -lSR | sort -k 5 -n
Or, if you want to display only file names:
ls -lSR | sort -k 5 -n | sed 's/ \+/\t/g' | cut -f 9
Parsing ls or variations thereof will add unnecesary complexity.
Sorted file paths by file size:
find src -type f -printf '%s\t%p\n' | sort -n | cut -f2-
Notes:
Change sort -n to sort -nr to get reverse order
The question had -print0 but catering for file names containing newlines seem pedantic.
The question mentioned relative paths, and changing %p to %P will get relative paths under src/
find -type f -exec du -sm {} \; | sort -nk1
Size in MiB, path is relative.
find /folder -type f -exec ls -S {} +
WARNING! I use not -exec ... \;, but -exec ... {} +. This construction doesn't pluck every file just it was found, but it waits, while all files will be found, and then puts them to one command (this time — ls) as one big list of arguments.
Then ls is looking at files and, because key -S, sorts them by size, largest first.
Literally none of these answers actually worked.
Here's what I made.
#!/bin/bash
################# ls-by-min ################
## List By Min File Size ##
## Copyright (c) 2020 Theodore R. Smith ##
## License: MIT ##
if [ -z "$1" ]; then
echo "Error: You must specify a minimum file size (in MB)."
exit 1
fi
FILE_SIZE=$1
if [ "$2" = "-r" ]; then
MAXDEPTH=512
else
MAXDEPTH=1
fi
find . -maxdepth ${MAXDEPTH} -type f -size "+${FILE_SIZE}M" -exec du -sm {} \; | sort -rnk1 | sed 's/^[0-9]\+\t*//g'
https://github.com/hopeseekr/BashScripts/blob/master/ls-by-min
You run it by doing:
ls-by-min 100 [-r]
and it will list, ordered biggest to smallest, only files that are 100 MB or bigger in the current directory. Pass in -r for it to be recursive.

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