Revisions to How do I get the size of a directory on the command line?

deleted 98 characters in body

edited Feb 22, 2020 at 0:26

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Here is a POSIX script that will work with:

A file
Files
A directory
Directories

#!/bin/sh ls -ARgo "$@" | awk '{q += $3} END {print q}'

Source

ls -A -R -g -o "$@" | awk '{n1 += $3} END {print n1}'

Here is a POSIX script that will work with:

A file
Files
A directory
Directories

#!/bin/sh ls -ARgo "$@" | awk '{q += $3} END {print q}'

Source

Here is a POSIX script that will work with:

A file
Files
A directory
Directories

ls -A -R -g -o "$@" | awk '{n1 += $3} END {print n1}'

deleted 11 characters in body

Source Link

edited Feb 9, 2018 at 1:52

Zombo

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Here is a POSIX functionscript that will work with:

A file
Files
A directory
Directories

file_size() {#!/bin/sh  ls -ARnARgo "$@" | awk '{zq += $5$3} END {print zq}' }

Source Source

Here is a POSIX function that will work with:

A file
Files
A directory
Directories

file_size() {  ls -ARn "$@" | awk '{z += $5} END {print z}' }

Source

Here is a POSIX script that will work with:

A file
Files
A directory
Directories

#!/bin/sh ls -ARgo "$@" | awk '{q += $3} END {print q}'

Source

added 1 character in body

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edited Apr 10, 2017 at 18:33

Zombo

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Here is a POSIX function that will work with:

A file
Files
A directory
Directories

std_sizefile_size() { ls -ARn "$@" | awk '{z += $5} END {print z}' }

Source Source

Here is a POSIX function that will work with:

A file
Files
A directory
Directories

std_size() { ls -ARn "$@" | awk '{z += $5} END {print z}' }

Source

Here is a POSIX function that will work with:

A file
Files
A directory
Directories

file_size() { ls -ARn "$@" | awk '{z += $5} END {print z}' }

Source

added 12 characters in body

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edited Mar 28, 2017 at 0:04

Zombo

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added 22 characters in body

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edited Mar 10, 2017 at 19:20

Zombo

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answered Mar 10, 2017 at 5:50

Zombo

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