Printing the sum-size of copied files in bash

up vote
6
down vote

favorite

I would like from the code below to print the sum-size of the files being copied instead of the size of each copied file separately.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

for i in *.tiff; do
 size=$(wc -c < $i) 
 cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"
done

This bash prints every second the size of the file being copied.

For example, in second 1 it copies file A and prints the size of file A. In second 2 it copies file B and prints the size of file B instead of the total size of files A+B. In second 3 it copies the file C and prints the size of file C instead of the total size of files A+B+C

How can I do it in bash specifically?

edited Sep 9 at 16:11

asked Sep 9 at 16:07

TNT

308111

1

You should note that it doesn't print the size every second, it sleeps one second between echo file. If it takes long enough that you want to display some progress, you probably don't want an extra delay.
â€“Â RalfFriedl
Sep 9 at 16:31

@ RalfFriedl. Thanks! probably just sleep 0.1 ?
â€“Â Goro
Sep 9 at 18:21

add a commentÂ |Â

up vote
6
down vote

favorite

I would like from the code below to print the sum-size of the files being copied instead of the size of each copied file separately.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

for i in *.tiff; do
 size=$(wc -c < $i) 
 cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"
done

This bash prints every second the size of the file being copied.

How can I do it in bash specifically?

edited Sep 9 at 16:11

asked Sep 9 at 16:07

TNT

308111

1

You should note that it doesn't print the size every second, it sleeps one second between echo file. If it takes long enough that you want to display some progress, you probably don't want an extra delay.
â€“Â RalfFriedl
Sep 9 at 16:31

@ RalfFriedl. Thanks! probably just sleep 0.1 ?
â€“Â Goro
Sep 9 at 18:21

add a commentÂ |Â

up vote
6
down vote

favorite

I would like from the code below to print the sum-size of the files being copied instead of the size of each copied file separately.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

for i in *.tiff; do
 size=$(wc -c < $i) 
 cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"
done

This bash prints every second the size of the file being copied.

How can I do it in bash specifically?

edited Sep 9 at 16:11

asked Sep 9 at 16:07

TNT

308111

I would like from the code below to print the sum-size of the files being copied instead of the size of each copied file separately.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

for i in *.tiff; do
 size=$(wc -c < $i) 
 cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"
done

This bash prints every second the size of the file being copied.

How can I do it in bash specifically?

bash

edited Sep 9 at 16:11

asked Sep 9 at 16:07

TNT

308111

edited Sep 9 at 16:11

asked Sep 9 at 16:07

TNT

308111

edited Sep 9 at 16:11

asked Sep 9 at 16:07

TNT

308111

asked Sep 9 at 16:07

TNT

308111

asked Sep 9 at 16:07

TNT

308111

1

You should note that it doesn't print the size every second, it sleeps one second between echo file. If it takes long enough that you want to display some progress, you probably don't want an extra delay.
â€“Â RalfFriedl
Sep 9 at 16:31

@ RalfFriedl. Thanks! probably just sleep 0.1 ?
â€“Â Goro
Sep 9 at 18:21

add a commentÂ |Â

1

You should note that it doesn't print the size every second, it sleeps one second between echo file. If it takes long enough that you want to display some progress, you probably don't want an extra delay.
â€“Â RalfFriedl
Sep 9 at 16:31

@ RalfFriedl. Thanks! probably just sleep 0.1 ?
â€“Â Goro
Sep 9 at 18:21

You should note that it doesn't print the size every second, it sleeps one second between echo file. If it takes long enough that you want to display some progress, you probably don't want an extra delay.
â€“Â RalfFriedl
Sep 9 at 16:31

@ RalfFriedl. Thanks! probably just sleep 0.1 ?
â€“Â Goro
Sep 9 at 18:21

add a commentÂ |Â

4 Answers
4

active

oldest

votes

up vote
5
down vote

Just sum up the single sizes and echo it afterwards.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

size_sum=0 
for i in *.tiff; do
 size=$(wc -c < $i) 
 size_sum=$((size_sum + size))
 cp -v $i $out | sleep 1 
done

echo $size_sum

answered Sep 9 at 16:23

Thomas

3,62141124

add a commentÂ |Â

up vote
1
down vote

size=$(wc -c < $i)

This isn't a very smart way to get the size of a file. Running wc like that will involve reading the whole file, which really is completely unnecessary since all you want is the size which the filesystem already knows and can tell you.

This is mitigated by the fact that you copy the file afterwards, and so need to read it in any case. A moderately-sized file will probably stay in the cache for the duration, so the reads don't actually hit the disk. But still, it would seem sensible to just read the size.

With GNU utilities, you can get the size of the file with stat -c %s "$filename". Sadly, the options stat takes are different on other systems, and I don't think there's a better, more portable way (apart from running Perl)

cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"

Here, the pipeline seems odd. cp doesn't produce any output (to standard output), at least not without -i, so there's nothing to pipe. Even if it did, sleep doesn't read anything so any output from cp would be wasted. The same thing happens between sleep and echo. This is basically the total opposite of what pipelines are usually used for, but it does have the effect that all the commands run in parallel, and the shell waits for all of them to complete, so that pipeline takes at least one second to run, regardless of how fast the copy would otherwise be.

If you want to sum the file sizes, the shell's arithmetic expansion $(( .. )) is what you probably want. So, without any extra sleeps:

dest=/path/to/destination
total=0
for file in *.tiff; do
 size=$(stat -c %s "$file");
 printf "current %d total %dn" "$size" "$((total += size))"
 cp -- "$file" "$dest"
done

answered 2 days ago

ilkkachu

50.8k678140

add a commentÂ |Â

up vote
0
down vote

To save on the arithmetic in your script, you can just take the size of the concatenation of the files:

cat *.tiff | wc

The UN*X kernel will conspire with you to make this just as efficient as taking the sizes of the files separately.

edited 2 days ago

muru

33.6k577144

answered 2 days ago

Rob Arthan

101

New contributor

I think this should be wc -c.
â€“Â user1934428
2 days ago

add a commentÂ |Â

up vote
0
down vote

It is possible to sum with += if the variable is declared as integer:

$ declare -i size
$ size+=100; echo "$size"
100
$ size+=200; echo "$size"
300

Your script (with variable expansions correctly quoted):

#!/bin/bash
input_folder=./a/b/c
output_folder=./d/e/f
declare -i size

for i in "$input_folder"/*; do
 size+=$(wc -c < "$i") 
 cp -v "$i" "$output_folder" | sleep 1 | echo -ne "$size33[0Kr"
done

answered 2 days ago

Isaac

6,9221834

add a commentÂ |Â

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "106"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2funix.stackexchange.com%2fquestions%2f467876%2fprinting-the-sum-size-of-copied-files-in-bash%23new-answer', 'question_page');

);

Post as a guest

Name

4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

up vote
5
down vote

Just sum up the single sizes and echo it afterwards.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

size_sum=0 
for i in *.tiff; do
 size=$(wc -c < $i) 
 size_sum=$((size_sum + size))
 cp -v $i $out | sleep 1 
done

echo $size_sum

answered Sep 9 at 16:23

Thomas

3,62141124

add a commentÂ |Â

up vote
5
down vote

Just sum up the single sizes and echo it afterwards.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

size_sum=0 
for i in *.tiff; do
 size=$(wc -c < $i) 
 size_sum=$((size_sum + size))
 cp -v $i $out | sleep 1 
done

echo $size_sum

answered Sep 9 at 16:23

Thomas

3,62141124

add a commentÂ |Â

up vote
5
down vote

Just sum up the single sizes and echo it afterwards.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

size_sum=0 
for i in *.tiff; do
 size=$(wc -c < $i) 
 size_sum=$((size_sum + size))
 cp -v $i $out | sleep 1 
done

echo $size_sum

answered Sep 9 at 16:23

Thomas

3,62141124

Just sum up the single sizes and echo it afterwards.

#!/bin/bash
input_folder=/a/b/c
output_folder=/d/e/f

cd $input_folder

size_sum=0 
for i in *.tiff; do
 size=$(wc -c < $i) 
 size_sum=$((size_sum + size))
 cp -v $i $out | sleep 1 
done

echo $size_sum

answered Sep 9 at 16:23

Thomas

3,62141124

answered Sep 9 at 16:23

Thomas

3,62141124

answered Sep 9 at 16:23

Thomas

3,62141124

answered Sep 9 at 16:23

Thomas

3,62141124

add a commentÂ |Â

up vote
1
down vote

size=$(wc -c < $i)

cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"

If you want to sum the file sizes, the shell's arithmetic expansion $(( .. )) is what you probably want. So, without any extra sleeps:

dest=/path/to/destination
total=0
for file in *.tiff; do
 size=$(stat -c %s "$file");
 printf "current %d total %dn" "$size" "$((total += size))"
 cp -- "$file" "$dest"
done

answered 2 days ago

ilkkachu

50.8k678140

add a commentÂ |Â

up vote
1
down vote

size=$(wc -c < $i)

cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"

If you want to sum the file sizes, the shell's arithmetic expansion $(( .. )) is what you probably want. So, without any extra sleeps:

dest=/path/to/destination
total=0
for file in *.tiff; do
 size=$(stat -c %s "$file");
 printf "current %d total %dn" "$size" "$((total += size))"
 cp -- "$file" "$dest"
done

answered 2 days ago

ilkkachu

50.8k678140

add a commentÂ |Â

up vote
1
down vote

size=$(wc -c < $i)

cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"

If you want to sum the file sizes, the shell's arithmetic expansion $(( .. )) is what you probably want. So, without any extra sleeps:

dest=/path/to/destination
total=0
for file in *.tiff; do
 size=$(stat -c %s "$file");
 printf "current %d total %dn" "$size" "$((total += size))"
 cp -- "$file" "$dest"
done

answered 2 days ago

ilkkachu

50.8k678140

size=$(wc -c < $i)

cp -v $i $out | sleep 1 | echo -ne "$size%33[0Kr"

If you want to sum the file sizes, the shell's arithmetic expansion $(( .. )) is what you probably want. So, without any extra sleeps:

dest=/path/to/destination
total=0
for file in *.tiff; do
 size=$(stat -c %s "$file");
 printf "current %d total %dn" "$size" "$((total += size))"
 cp -- "$file" "$dest"
done

answered 2 days ago

ilkkachu

50.8k678140

answered 2 days ago

ilkkachu

50.8k678140

answered 2 days ago

ilkkachu

50.8k678140

answered 2 days ago

ilkkachu

50.8k678140

add a commentÂ |Â

up vote
0
down vote

To save on the arithmetic in your script, you can just take the size of the concatenation of the files:

cat *.tiff | wc

The UN*X kernel will conspire with you to make this just as efficient as taking the sizes of the files separately.

edited 2 days ago

muru

33.6k577144

answered 2 days ago

Rob Arthan

101

New contributor

I think this should be wc -c.
â€“Â user1934428
2 days ago

add a commentÂ |Â

up vote
0
down vote

To save on the arithmetic in your script, you can just take the size of the concatenation of the files:

cat *.tiff | wc

The UN*X kernel will conspire with you to make this just as efficient as taking the sizes of the files separately.

edited 2 days ago

muru

33.6k577144

answered 2 days ago

Rob Arthan

101

New contributor

I think this should be wc -c.
â€“Â user1934428
2 days ago

add a commentÂ |Â

up vote
0
down vote

To save on the arithmetic in your script, you can just take the size of the concatenation of the files:

cat *.tiff | wc

The UN*X kernel will conspire with you to make this just as efficient as taking the sizes of the files separately.

edited 2 days ago

muru

33.6k577144

answered 2 days ago

Rob Arthan

101

New contributor

To save on the arithmetic in your script, you can just take the size of the concatenation of the files:

cat *.tiff | wc

The UN*X kernel will conspire with you to make this just as efficient as taking the sizes of the files separately.

edited 2 days ago

muru

33.6k577144

answered 2 days ago

Rob Arthan

101

New contributor

edited 2 days ago

muru

33.6k577144

edited 2 days ago

muru

33.6k577144

edited 2 days ago

muru

33.6k577144

answered 2 days ago

Rob Arthan

101

New contributor

answered 2 days ago

Rob Arthan

101

answered 2 days ago

Rob Arthan

101

New contributor

Rob Arthan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

I think this should be wc -c.
â€“Â user1934428
2 days ago

add a commentÂ |Â

I think this should be wc -c.
â€“Â user1934428
2 days ago

I think this should be wc -c.
â€“Â user1934428
2 days ago

add a commentÂ |Â

up vote
0
down vote

It is possible to sum with += if the variable is declared as integer:

$ declare -i size
$ size+=100; echo "$size"
100
$ size+=200; echo "$size"
300

Your script (with variable expansions correctly quoted):

#!/bin/bash
input_folder=./a/b/c
output_folder=./d/e/f
declare -i size

for i in "$input_folder"/*; do
 size+=$(wc -c < "$i") 
 cp -v "$i" "$output_folder" | sleep 1 | echo -ne "$size33[0Kr"
done

answered 2 days ago

Isaac

6,9221834

add a commentÂ |Â

up vote
0
down vote

It is possible to sum with += if the variable is declared as integer:

$ declare -i size
$ size+=100; echo "$size"
100
$ size+=200; echo "$size"
300

Your script (with variable expansions correctly quoted):

#!/bin/bash
input_folder=./a/b/c
output_folder=./d/e/f
declare -i size

for i in "$input_folder"/*; do
 size+=$(wc -c < "$i") 
 cp -v "$i" "$output_folder" | sleep 1 | echo -ne "$size33[0Kr"
done

answered 2 days ago

Isaac

6,9221834

add a commentÂ |Â

up vote
0
down vote

It is possible to sum with += if the variable is declared as integer:

$ declare -i size
$ size+=100; echo "$size"
100
$ size+=200; echo "$size"
300

Your script (with variable expansions correctly quoted):

#!/bin/bash
input_folder=./a/b/c
output_folder=./d/e/f
declare -i size

for i in "$input_folder"/*; do
 size+=$(wc -c < "$i") 
 cp -v "$i" "$output_folder" | sleep 1 | echo -ne "$size33[0Kr"
done

answered 2 days ago

Isaac

6,9221834

It is possible to sum with += if the variable is declared as integer:

$ declare -i size
$ size+=100; echo "$size"
100
$ size+=200; echo "$size"
300

Your script (with variable expansions correctly quoted):

#!/bin/bash
input_folder=./a/b/c
output_folder=./d/e/f
declare -i size

for i in "$input_folder"/*; do
 size+=$(wc -c < "$i") 
 cp -v "$i" "$output_folder" | sleep 1 | echo -ne "$size33[0Kr"
done

answered 2 days ago

Isaac

6,9221834

answered 2 days ago

Isaac

6,9221834

answered 2 days ago

Isaac

6,9221834

answered 2 days ago

Isaac

6,9221834

add a commentÂ |Â

draft saved

draft discarded

draft saved

draft discarded

Post as a guest

Name

Search This Blog

Iyfjky