And I started playing with dmenu and it seems such an automation for almost every thing. Unfortunately I'm not familiar with bash and it should be on my list.
I have a folder for my markdowns with subfolders containing my files. I'm trying to have a script to show them in dmenu while using an alias.
If the path to a file is
/home/user/docs/markdown/practice01/rmd/network.rmd
I would like to have
network
as an option in my dmenu. So when I choose
network -----> /home/user/docs/markdown/practice01/rmd/network.rmd
Here is my broken script. There are a few things I'm missing.
This way I get full path on my dmenu which i don't need. I tried to read about associative arrays but I can't figure it out in bash.
This script works but in case I decide to ESC and exit, still it opens up an empty vim in my directory. Hence, I should know if statements huh!
#!/bin/bash
DMenu=("dmenu -l 10 -i -nb "#eaeaea" -sb "#E53935" -nf "#474747"")
cd ~/docs/markdown/
target=$(find -type f -name '*.rmd' | $DMenu)
st vim "$target"
I made a little example. But the problem is that it is a manual work to add each file, which definitely we don't wanna do right!
#!/bin/bash
declare -A dotfiles
dotfiles[i3]="/home/user/dotfiles/i3/.config/i3/config"
dotfiles[vimrc]="/home/user/dotfiles/vim/.vimrc"
list=("i3\nvimrc")
target=$(echo -e $list | dmenu -i -nb "#eaeaea" -sb "#E53935" -nf "#474747")
st vim "${dotfiles["$target"]}"
Thank you
Associative arrays can be weird... but returning output to a variable makes it easier to manipulate as any other string in bash, as shown in the example below:
prefix="$HOME/git/notes"
suffix=".md"
shopt -s nullglob globstar
item=( "$prefix"/**/*${suffix}) # Search *.md in all dirs/subdirs
item=( "${item[#]#"$prefix"/}" )
item=( "${item[#]%${suffix}}" ) # Removes '.md' string from item name
result=$(printf '%s\n' "${item[#]}" | dmenu)
[[ -n $result ]] || exit # exit if nothing is found
gedit "${prefix}/${result}.md" # Open file by adding again '.md'
When the percent sign (%) is used in the pattern ${variable%substring}, it will return content of the variable with the shortest occurrence of substring deleted from the back of the variable.
Listed below for reference are 2 examples I wrote, one in Bash and the other in Python, for managing pass and markdown notes with dmenu:
dmenu-pass.sh
dmenu-launch.py
Also, listed below are a couple nice articles that might help you out:
The weird, wondrous world of Bash arrays
Advanced Bash-Scripting Guide: Manipulating Strings
Instead of putting some code in an array, use a function!
my_dmenu() {
dmenu -l 10 -i -nb "#eaeaea" -sb "#e53935" -nf "#474747"
}
If your markdown files are all in the same folder (and not in subfolders), you certainly don't need find: use a glob instead! and if your files are in subfolders, use a glob instead (with the globstar shell option).
All in all:
#!/bin/bash
my_dmenu() {
dmenu -l 10 -i -nb "#eaeaea" -sb "#e53935" -nf "#474747"
}
base_dir=~/docs/markdown
# Also, check the return code of cd!
cd "$base_dir" || { echo >&2 "Can't cd to $base_dir. Exiting"; exit 1; }
# Using a glob: use the shell option nullglob
shopt -s nullglob
files=( *.rmd )
# Check that there are some files found:
if (( ${#files[#]} == 0 )); then
echo "No files found. Exiting."
exit 1
fi
# Now we're ready to send the files to dmenu:
chosen_file=$(printf '%s\n' "${files[#]}" | my_dmenu)
# If dmenu returns nothing: don't launch vim!
if [[ ! $chosen_file ]]; then
echo "No files selected. Exiting."
exit 1
fi
# Now you can launch vim!
st vim "$chosen_file"
If you also want to find the *.rmd files in subfolders: use instead:
shopt -s nullglob globstar
files=( **/*.rmd )
Edit to address the requirement in your comment (and the edit of your question):
If you want to strip the .rmd suffix to show in dmenu, use:
chosen_file=$(printf '%s\n' "${files[#]%.rmd}" | my_dmenu)
# ...
st vim "$chosen_file.rmd"
The expansion ${files[#]%.rmd} will strip the suffix .rmd from each field of the array files. Don't forget to add this suffix back when you edit the file (as shown in the last line).
dmenuoptions="-l 10 -i -nb '#eaeaea' -sb '#E53935' -nf '#474747'"
st -e vim $(find ~/docs/markdown -type f -name '*.rmd' | dmenu $dmenuoptions)
Related
This question already has answers here:
Run script on multiple files
(3 answers)
Closed 3 years ago.
I'm very new to bash. I have ten text files that I want to edit with the same line of code.
#!/bin/bash
sed -i -e 's/.\{6\}/&\n/g' -e 's/edit/edit2/g' | tr -d "\n" | sed 's/edit2/edit/g'| grep -o "here.*there" | sed -r '/^.{,100}$/d'
< files 1-10
I know I could use sed -f sed.sh <file1 >file1 but that only works with sed commands and it only works one file at a time?
Do I have to run a loop?
There's some great existing answers on the Unix stack exchange that help deal with your problem. Specifically, from this post, they use a loop to recursively loop through all the files in a particular directory, as follows:
( shopt -s globstar dotglob;
for file in **; do
if [[ -f $file ]] && [[ -w $file ]]; then
sed -i -- 's/foo/bar/g' "$file"
fi
done
)
Note the line, shopt -s globstar dotglob;, which allows us to use globbing patterns in the for loop. We also enclose the code in brackets, to prevent the shopt -s globstar dotglob; line option from becoming a global setting.
If you would like to apply this example to your file, you can just place your files in the current directory, and the code would probably look something like this:
( shopt -s globstar dotglob;
for file in **; do
if [[ -f $file ]] && [[ -w $file ]]; then
sed -i -e 's/.\{6\}/&\n/g' -e 's/edit/edit2/g' | tr -d "\n" | sed 's/edit2/edit/g' | grep -o "here.*there" | sed -r '/^.{,100}$/d' "$file"
fi
done
)
Note that we have placed a "$file" variable beside each of the seds that you used in your code, this replaces the name of the file for each command.
There is another example given in the code that allows you to pick which files to run on, rather than all the files in a directory, which you can also re-purpose for your code, as given here:
( shopt -s globstar dotglob
sed -i -- 's/foo/bar/g' **baz*
sed -i -- 's/foo/bar/g' **.baz
)
To answer your question of doing a loop on each line, you will need to put a loop for each line inside your for loop, like so:
while read line ; do
: sed -i -e 's/.\{6\}/&\n/g' -e 's/edit/edit2/g' | tr -d "\n" | sed 's/edit2/edit/g' | grep -o "here.*there" | sed -r '/^.{,100}$/d' "$line”
done
)
Although the for loop can be useful for dealing with files in recursive directories, I would recommend against also using another loop to grab lines, since it muddies your code, and it’s possible there is a better way to do it without parsing line by line.
The linked question is a fairly complete guide to many of the cases you may come across, and is also worth a read if you want to learn more.
Hope that helps!
You could use a for loop.
You could use the tool parallel.
Example
Create a set of test files using a for-loop
mkdir -p /tmp/so58333536
cd /tmp/so58333536
for i in 1.txt 2.txt 3.txt 4.txt 5.txt;do echo "The answer is 41" > $i;done
cat /tmp/so58333536/*
Now correct your mistake using parallel [1].
mkdir /tmp/so58333536.new
ls /tmp/so58333536/* |parallel "sed 's/41/42/' {} > /tmp/so58333536.new/{/}"
cat /tmp/so58333536.new/*
{}:: refers to the current file
{/}:: refers to name of the current file (path is removed)
Reads: List all files in so58333536 and apply the following sed command to each file and write the output to so58333536.new.
[1] Another option is to use sed -i for in-place editing.
Be very carefull with this!! Mistakes can cause serious damages!
# !! Do not use -i option regularly !!
ls /tmp/so58333536/* |parallel "sed -i 's/41/42/'"
I am trying to move files from a directory if it is not empty. This script is added in cronjob but it is always executing regardless of files are present or not? What is wrong in this thing?
#!/bin/bash
logFolder="/dstDir/`date '+%Y-%m-%d/%H-%M'`"
tempLogFolder="sourceDir"
if [ -z "$(ls -A $tempLogFolder | grep *.log)" ]; then
mkdir -p $logFolder
mv $tempLogFolder/*.log $logFolder/
fi
Don't parse the output of ls. You don't need any external commands to count files in a directory. The shell can do it itself.
Try creating a helper function that checks how many arguments are passed to it. Then have the shell expand the glob "$tempLogFolder"/*.log. No need for ls or grep. The only trick is enabling the nullglob option so if no files exist the glob expands to nothing.
files_exist() { (($# > 0)); }
shopt -s nullglob
if files_exist "$tempLogFolder"/*.log; then
mkdir -p "$logFolder"
mv "$tempLogFolder"/*.log "$logFolder"/
fi
grep accepts regex, not glob. If you want to use glob like *.log you put it in your ls something like ls .... *.log (using ls output is not 100% safe).
If you want to use grep, please use regex, I guess what you meant is \.log$.
If you want to check if grep has found matches, you should check the return code ($?) of grep, instead of using -z test.
If you put it in crontab, the script will always be executed.
I think you want,
if [ -n "$(ls -A $tempLogFolder | grep *.log)" ]; then
...
because you are only interested in whether the string returned has a non-zero length.
This question already has answers here:
How to skip the for loop when there are no matching files?
(2 answers)
Closed 3 years ago.
I'm trying to iterate through all zip files in a bash script (I'm using Cygwin, but I kind of doubt this is a bug with Cygwin).
It looks like this right now:
for z in *.zip
do
echo $z
done
which works well when there are zip files in the folder, and it echos exactly the zip files and nothing but the zip files. However, when I do it on a folder that's empty, it echos *.zip, when I'd rather it echo nothing.
What should I be doing? I don't think the right solution is if [ $z != "*.zip ]... but is it?
This is the expected behavior. From the documentation:
If no matching filenames are found, and the shell option nullglob is disabled, the word is left unchanged. If the nullglob option is set, and no matches are found, the word is removed.
So the solution is to set the nullglob option before the loop:
shopt -s nullglob
for z in *.zip
...
As one of its steps for executing a command, the shell may perform path expansion in which case it will replace a string, such as *.zip with the list of files that match that glob. If there are no such files, then the string is left unchanged. A reasonable solution is:
for z in *.zip
do
[ -f "$z" ] && echo $z
done
[ -f "$z" ] verifies that the file exists and is a regular file. The && means that echo will be executed only if that test is passed.
Either that or turning on the nullglob option.
$ set -x
$ echo *.none
+ echo '*.none'
*.none
$ shopt -s nullglob
+ shopt -s nullglob
$ echo *.none
$ echo *.none
+ echo
I have a directory with lots of files. I want to keep only the 6 newest. I guess I can look at their creation date and run rm on all those that are too old, but is the a better way for doing this? Maybe some linux command I could use?
Thanks!
:)
rm -v $(ls -t mysvc-*.log | tail -n +7)
ls -t, list sorted by time
tail -n +7, +7 here means length-7, so all but first 7 lines
$() makes a list of strings from the enclosed command output
rm to remove the files, of course
Beware files with space in their names, $() splits on any white-space!
Here's my take on it, as a script. It does handle spaces in file names even if it is a bit of a hack.
#!/bin/bash
eval set -- $(ls -t1 | sed -e 's/.*/"&"/')
if [[ $# -gt 6 ]] ; then
shift 6
while [[ $# -gt 0 ]] ; do
echo "remove this file: $1" # rm "$1"
shift
done
fi
The second option to ls up there is a "one" for one file name per line. Doesn't actually seem to matter, though, since that appears to be the default when ls isn't feeding a tty.
I'm writing a script to batch resize images. Originally I was applying an operation for file in $(ls $1), but I would like to be able to use globbing, so I'm looking at something more like for file in $(echo $1). The problem is that dotglob may or may not be enabled, so echo * could return hidden files (notably, .DS_Store), which cause convert to throw an error and stop the script. I would like the default behavior of the command to be that if I cd into a directory full of images and execute resize * 400x400 jpg, all of the images will be resized excluding hidden files, regardless of whether dotglob is enabled.
So, in pseudo code, I'm looking for:
for file in $(echo $1 | [filter-hidden-files])
Here is my script with the older behavior. Will update with new behavior when I find a solution:
# !/bin/bash
# resize [folder] [sizeXxsizeY] [outputformat]
# if [outputformat] is omitted, the input file format is assumed
for file in $(ls $1)
do
IMGNAME=$(echo "$file" | cut -d'.' -f1)
if test -z $3
then
EXTENSION=$(echo "$file" | cut -d'.' -f2)
convert $1/$file -resize $2 -quality 100 $1/$IMGNAME-$2.$EXTENSION
echo "$file => $IMGNAME-$2.$EXTENSION"
else
convert $1/$file -resize $2 -quality 100 $1/$IMGNAME-$2.$3
echo "$file => $IMGNAME-$2.$3"
fi
done
Here is the current script:
# !/bin/bash
# resize [pattern] [sizeXxsizeY] [outputformat]
# if [outputformat] is omitted, the input file format is assumed
for file in $(echo $1)
do
IMGNAME=$(echo "$file" | cut -d'.' -f1)
if test -z $3 && if test -f $3
then
EXTENSION=$(echo "$file" | cut -d'.' -f2)
convert $file -resize $2 -quality 100 $IMGNAME-$2.$EXTENSION
echo "$file => $IMGNAME-$2.$EXTENSION"
else
convert $file -resize $2 -quality 100 $IMGNAME-$2.$3
echo "$file => $IMGNAME-$2.$3"
fi
done
Given the command resize * 400x400, convert throws an error as it cannot process .DS_Store (a hidden file residing in every file on an OSX system). As I will never be processing hidden images, I would like to automatically filter them. I've been trying to do this with grep or find, but I haven't figured it out yet.
New script goes here:
for file in $(echo $1)
do
I would suggest changing the commandline of your script from resize * 400x400 to resize 400x400 *, the script would be more like the standard unix tools that way (grep, sed, ln, etc.). You would not have to unset the dotglob option in your script (which is better since it's up to the user of the script if he wants hidden files globbed or not).
Your script would look something like this:
#!/bin/bash
OUTPUTFORMAT=$1
# Remove original $1 from the list of arguments
shift
for i in "$#"
do
# Use $OUTPUTFORMAT here
etc....
If you do not want to change the commandline for your script. You could try setting GLOBIGNORE
export GLOBIGNORE=".*"
Or if extglob is set you could try file globbing like so:
echo !(.*)
There is a dotglob shell option that decides if files starting with . are included when globbing. You can check if this is the case with
shopt dotglob
You also can explicitly disable it in your script:
#!/bin/bash
shopt -u dotglob
for file in $1/*; do
...
done
there's no need to use ls with a for loop, most of the time its useless. also the for loop with * doesn't return hidden files, unless you specifically specify it. To show hidden files,
for files in .*
do
echo $files
done
as you are getting a list of files in $* you can check them one by one
for i in $*
do
expr $i : '^\..*' > /dev/null && continue
# process file
done