I want to display all the column headers when I type ls -l command in bash shell in unix/linux
When we type ls -ltr on command prompt we get something like the following.
-r--r--r-- 2 makerpm root 1898 Jan 28 14:52 sample3
-r--r--r-- 2 makerpm root 1898 Jan 28 14:52 sample1
What I want is to know whether ls has any options to display with column headers:
File_Permissions Owner Group Size Modified_Time Name
-r--r--r-- 2 makerpm root 1898 Jan 28 14:52 sample3
-r--r--r-- 2 makerpm root 1898 Jan 28 14:52 sample1
exa is a replacement/enhancement for ls. If you pass on the arguments -lh with exa, it will include a header row printing the column names like so:
exa -lh
Example output:
Permissions Size User Date Modified Name
.rwx------ 19 username 29 Sep 11:25 dont_cra.sh
drw-r----- - username 29 Sep 11:26 f1
.rw-r--r--# 811k username 29 Sep 11:25 row_count.dat
.rw-r--r-- 54 username 29 Sep 11:25 some_text.txt
You can set up an alias in .bashrc that replaces ls with exa.
The last answer using sed was slick, but unfortunately, if you have color added to your output (which most people do) it removes all color. I would like to suggest a better way, and ironically, simpler too.
First off, my .bashrc USED to have the following:
# enable color support of ls and also add handy aliases
if [ -x /usr/bin/dircolors ]; then
test -r ~/.dircolors && eval "$(dircolors -b ~/.dircolors)" || eval "$(dircolors -b)"
alias ls='ls --color=auto'
fi
alias ls='ls -AFhls --color --group-directories-first'
To be honest, you don't need the dircolors part, that is just a little extra I use, you could have something as simple as:
alias ls='ls --color=auto'
I wanted column headers too, Googled it, and wound up here. However after I tried what the previous user suggested, with sed and realizing everything was white, and my colors have all gone away.
That's when I tried something different in my .bashrc file, and it worked.
Simply alias ls, echo first, then place a semi-colon, then your ls command.
My .bashrc file now has the following line.
alias ls='echo "Dir Size|Perms|Link Count|Owner|Group|Size|Mod. Time|Name"; ls -AFhls --color --group-directories-first'
When doing it this way, utilizing echo instead of sed, all colors continue to work.
Not sure if this is specific to my terminal's output but, this worked for me.
And this is the output it yields. As you can see, it preserves color using this method. Just be sure to keep the ${1} variable in double quotes so files and directories with spaces in the name won't cause an error.
Here is the code so you can copy and paste for testing.
long_ls() {
local VAR="Permissions|Owner|Group|Size|Modified|Name"
if [ ! "${1}" ]; then
echo -e "$VAR" | column -t -s"|" && ls -l
else
echo -e "$VAR" | column -t -s"|" && ls -l "${1}"
fi
}
alias lls=$"long_ls ${1}"
Related
I'd like to run some script.
It need alias to modify command.
but it seem invalid.
ex:
root#161310ea476b:/tmp# cat test.sh
#!/bin/bash
ls /tmp
root#161310ea476b:/tmp#
root#161310ea476b:/tmp# ./test.sh
test.sh
root#161310ea476b:/tmp# . test.sh
test.sh
now I used alias
root#161310ea476b:/tmp# shopt -s expand_aliases
root#161310ea476b:/tmp# alias ls="ls -al"
it works.
root#161310ea476b:/tmp# . test.sh
total 12
drwxrwxrwt 2 root root 4096 Jun 21 09:41 .
drwxr-xr-x 21 root root 4096 Jun 21 09:39 ..
-rwxr-xr-x 1 root root 22 Jun 21 09:41 test.sh
but this case wasn't works. alias seem invalid.
root#161310ea476b:/tmp# ./test.sh
test.sh
root#161310ea476b:/tmp#
How to fix this problem?
Thanks.
Aliases are not inherited by the subshell used to execute your script.
Please see the answer provided here:
Aliases in subshell
Alternatively, you could create your alias as a function (instead of an alias), export it, and THEN run your script.
Create a function...
function ls() {
/bin/ls -al
}
...export it...
export -f ls
...and run your script
. ./test.sh
Of course, don't forget you've created a function ls. (Note: it should only exist as long as the shell you exported it from --- and any subshells from it --- is/are open).
How do I calculate the sum total size of multiple files located in different directories?
I have a text file containing the full path and name of the files.
I figure a simple script using while read line and du -h might do the trick...
Example of text file (new2.txt) containing list of files to sum:
/mount/st4000/media/A/amediafile.ext
/mount/st4000/media/B/amediafile.ext
/mount/st4000/media/C/amediafile.ext
/mount/st4000/media/D/amediafile.ext
/mount/st4000/media/E/amediafile.ext
/mount/st4000/media/F/amediafile.ext
/mount/st4000/media/G/amediafile.ext
/mount/st4000/media/H/amediafile.ext
/mount/st4000/media/I/amediafile.ext
/mount/st4000/media/J/amediafile.ext
/mount/st4000/media/K/amediafile.ext
Note: the folder structure is not necessarily consecutive as in A..K
Based on the suggestion from AndreaT, adapting it slightly, I tried
while read mediafile;do du -b "$mediafile"|cut -f -1>>subtotals.txt;done<new2.txt
subtotals.txt looks like
733402685
944869798
730564608
213768
13332480
366983168
6122559750
539944960
735039488
1755005744
733478912
To add all the subtotals
sum=0; while read num; do ((sum += num)); done < subtotals.txt; echo $sum
Assuming that file input is like this
/home/administrator/filesum/cliprdr.c
/home/administrator/filesum/cliprdr.h
/home/administrator/filesum/event.c
/home/administrator/filesum/event.h
/home/administrator/filesum/main.c
/home/administrator/filesum/main.h
/home/administrator/filesum/utils.c
/home/administrator/filesum/utils.h
and the result of command ls -l is
-rw-r--r-- 1 administrator administrator 13452 Oct 4 17:56 cliprdr.c
-rw-r--r-- 1 administrator administrator 1240 Oct 4 17:56 cliprdr.h
-rw-r--r-- 1 administrator administrator 8141 Oct 4 17:56 event.c
-rw-r--r-- 1 administrator administrator 2164 Oct 4 17:56 event.h
-rw-r--r-- 1 administrator administrator 32403 Oct 4 17:56 main.c
-rw-r--r-- 1 administrator administrator 1074 Oct 4 17:56 main.h
-rw-r--r-- 1 administrator administrator 5452 Oct 4 17:56 utils.c
-rw-r--r-- 1 administrator administrator 1017 Oct 4 17:56 utils.h
the simplest command to run is:
cat filelist.txt | du -cb | tail -1 | cut -f -1
with following output (in bytes)
69370
Keep in mind that du prints actual disk usage rounded up to a multiple of (usually) 4kb instead of logical file size.
For small files this approximation may not be acceptable.
To sum one directory, you will have to do a while, and export the result to the parent shell.
I used an echo an the subsequent eval :
eval ' let sum=0$(
ls -l | tail -n +2 |\
while read perms link user uid size date day hour name ; do
echo -n "+$size" ;
done
)'
It produces a line, directly evaluated, which looks like
let sum=0+205+1201+1201+1530+128+99
You just have to reproduce twice this command on both folders.
The du command doesn't have a -b option on the unix systems I have available. And there are other ways to get file size.
Assuming you like the idea of a while loop in bash, the following might work:
#!/bin/bash
case "$(uname -s)" in
Linux) stat_opt=(-c '%s') ;;
*BSD|Darwin) stat_opt=(-f '%z') ;;
*) printf 'ERROR: I don'\''t know how to run on %s\n' "$(uname -s)" ;;
esac
declare -i total=0
declare -i count=0
declare filename
while read filename; do
[[ -f "$filename" ]] || continue
(( total+=$(stat "${stat_opt[#]}" "$filename") ))
(( count++ ))
done
printf 'Total: %d bytes in %d files.\n' "$total" "$count"
This would take your list of files as stdin. You can run it in BSD unix or in Linux -- the options to the stat command (which is not internal to bash) are the bit that are platform specific.
I have a directory in which I will have some folders with date format (YYYYMMDD) as shown below -
david#machineX:/database/batch/snapshot$ ls -lt
drwxr-xr-x 2 app kyte 86016 Oct 25 05:19 20141023
drwxr-xr-x 2 app kyte 73728 Oct 18 00:21 20141016
drwxr-xr-x 2 app kyte 73728 Oct 9 22:23 20141009
drwxr-xr-x 2 app kyte 81920 Oct 4 03:11 20141002
Now I need to extract latest date folder from the /database/batch/snapshot directory and then construct the command in my shell script like this -
./file_checker --directory /database/batch/snapshot/20141023/ --regex ".*.data" > shardfile_20141023.log
Below is my shell script -
#!/bin/bash
./file_checker --directory /database/batch/snapshot/20141023/ --regex ".*.data" > shardfile_20141023.log
# now I need to grep shardfile_20141023.log after above command is executed
How do I find the latest date folder and construct above command in a shell script?
Look, this is one of approaches, just grep only folders that have 8 digits:
ls -t1 | grep -P -e "\d{8}" | head -1
Or
ls -t1 | grep -E -e "[0-9]{8}" | head -1
You could try the following in your script:
pushd /database/batch/snapshot
LATESTDATE=`ls -d * | sort -n | tail -1`
popd
./file_checker --directory /database/batch/snapshot/${LATESTDATE}/ --regex ".*.data" > shardfile_${LATESTDATE}.log
See BashFAQ#099 aka "How can I get the newest (or oldest) file from a directory?".
That being said, if you don't care for actual modification time and just want to find the most recent directory based on name you can use an array and globbing (note: the sort order with globbing is subject to LC_COLLATE):
$ find
.
./20141002
./20141009
./20141016
./20141023
$ foo=( * )
$ echo "${foo[${#foo[#]}-1]}"
20141023
I am pretty stuck with that script.
#!/bin/bash
STARTDIR=$1
MNTDIR=/tmp/test/mnt
find $STARTDIR -type l |
while read file;
do
echo Found symlink file: $file
DIR=`sed 's|/\w*$||'`
MKDIR=${MNTDIR}${DIR}
mkdir -p $MKDIR
cp -L $file $MKDIR
done
I passing some directory to $1 parameter, this directory have three symbolic links. In while statement echoed only first match, after using sed I lost all other matches.
Look for output below:
[artyom#LBOX tmp]$ ls -lh /tmp/imp/
total 16K
lrwxrwxrwx 1 artyom adm 19 Aug 8 10:33 ok1 -> /tmp/imp/sym3/file1
lrwxrwxrwx 1 artyom adm 19 Aug 8 09:19 ok2 -> /tmp/imp/sym2/file2
lrwxrwxrwx 1 artyom adm 19 Aug 8 10:32 ok3 -> /tmp/imp/sym3/file3
[artyom#LBOX tmp]$ ./copy.sh /tmp/imp/
Found symlink file: /tmp/imp/ok1
[artyom#LBOX tmp]$
Can somebody help with that issue?
Thanks
You forgot to feed something to sed. Without explicit input, it reads nothing in this construction. I wouldn't use this approach anyway, but just use something like:
DIR=`dirname "$file"`
i have bunch of files that needs to be renamed.
file1.txt needs to be renamed to file1_file1.txt
file2.avi needs to be renamed to file2_file2.avi
as you can see i need the _ folowed by the original file name.
there are lot of these files.
So far all the answers given either:
Require some non-portable tool
Break horribly with filenames containing spaces or newlines
Is not recursive, i.e. does not descend into sub-directories
These two scripts solve all of those problems.
Bash 2.X/3.X
#!/bin/bash
while IFS= read -r -d $'\0' file; do
dirname="${file%/*}/"
basename="${file:${#dirname}}"
echo mv "$file" "$dirname${basename%.*}_$basename"
done < <(find . -type f -print0)
Bash 4.X
#!/bin/bash
shopt -s globstar
for file in ./**; do
if [[ -f "$file" ]]; then
dirname="${file%/*}/"
basename="${file:${#dirname}}"
echo mv "$file" "$dirname${basename%.*}_$basename"
fi
done
Be sure to remove the echo from whichever script you choose once you are satisfied with it's output and run it again
Edit
Fixed problem in previous version that did not properly handle path names.
For your specific case, you want to use mmv as follows:
pax> ll
total 0
drwxr-xr-x+ 2 allachan None 0 Dec 24 09:47 .
drwxrwxrwx+ 5 allachan None 0 Dec 24 09:39 ..
-rw-r--r-- 1 allachan None 0 Dec 24 09:39 file1.txt
-rw-r--r-- 1 allachan None 0 Dec 24 09:39 file2.avi
pax> mmv '*.*' '#1_#1.#2'
pax> ll
total 0
drwxr-xr-x+ 2 allachan None 0 Dec 24 09:47 .
drwxrwxrwx+ 5 allachan None 0 Dec 24 09:39 ..
-rw-r--r-- 1 allachan None 0 Dec 24 09:39 file1_file1.txt
-rw-r--r-- 1 allachan None 0 Dec 24 09:39 file2_file2.avi
You need to be aware that the wildcard matching is not greedy. That means that the file a.b.txt will be turned into a_a.b.txt, not a.b_a.b.txt.
The mmv program was installed as part of my CygWin but I had to
sudo apt-get install mmv
on my Ubuntu box to get it down. If it's not in you standard distribution, whatever package manager you're using will hopefully have it available.
If, for some reason, you're not permitted to install it, you'll have to use one of the other bash for-loop-type solutions shown in the other answers. I prefer the terseness of mmv myself but you may not have the option.
for file in file*.*
do
[ -f "$file" ] && echo mv "$file" "${file%%.*}_$file"
done
Idea for recursion
recurse() {
for file in "$1"/*;do
if [ -d "$file" ];then
recurse "$file"
else
# check for relevant files
# echo mv "$file" "${file%%.*}_$file"
fi
done
}
recurse /path/to/files
find . -type f | while read FN; do
BFN=$(basename "$FN")
NFN=${BFN%.*}_${BFN}
echo "$BFN -> $NFN"
mv "$FN" "$NFN"
done
I like the PERL cookbook's rename script for this. It may not be /bin/sh but you can do regular expression-like renames.
The /bin/sh method would be to use sed/cut/awk to alter each filename inside a for loop. If the directory is large you'd need to rely on xargs.
One should mention the mmv tool, which is especially made for this.
It's described here: http://tldp.org/LDP/GNU-Linux-Tools-Summary/html/mass-rename.html
...along with alternatives.
I use prename (perl based), which is included in various linux distributions. It works with regular expressions, so to say change all img_x.jpg to IMAGE_x.jpg you'd do
prename 's/img_/IMAGE_/' img*jpg
You can use the -n flag to preview changes without making any actual changes.
prename man entry
#!/bin/bash
# Don't do this like I did:
# files=`ls ${1}`
for file in *.*
do
if [ -f $file ];
then
newname=${file%%.*}_${file}
mv $file $newname
fi
done
This one won't rename sub directories, only regular files.