Increment the title of files output by a command in a shell script - linux

I made this simple bash script to take a full-screen screenshot and save it to the pictures folder
#!/usr/bin/bash
xfce4-screenshooter -f -s /home/rgcodes/Pictures/Screenshot_$scrshotcount.png
let "scrshotcount++"
...which runs into a problem. scrshotcount is a global variable I defined in /etc/environment to be incremented every time the script runs. However, the script fails to increment the variable globally, and causes the script to just overwrite the previous screenshot. Searches on Google and Stack Overflow revealed that the problem isn't straightforward at all (something about child shells being unable to change variables for parents), and finding some other method would be better.
Here's my question. How do we append numbers (in ascending order) to the screenshots the script throws out so that they are saved just like those taken on Windows?(Windows auto-suffixes matching filenames, rather than overwriting them, so all Screenshots have the same name 'Screenshot' and the number of times the screenshot command has been used.)
I am using #erikMD's method as a temporary stop-gap for now.

In addition to the excellent advice about using a date instead of a counter, here's a way to use a counter :/
dir=$HOME/Pictures
# find the current maximum value
current_max=$(
find "$dir" -name Screenshot_\*.png -print0 \
| sort -z -V \
| tail -z -n 1
)
if [[ ! $current_max =~ _([0-9]+)\.png ]]; then
echo "can't find the screenshot with the maximum counter value" >&2
exit 1
fi
# increment it
counter=$(( 1 + ${BASH_REMATCH[1]} ))
# and use it
xfce4-screenshooter -f -s "$dir/Screenshot_${counter}.png"
You'll have to manually create the Screenshot_1.png file.

#rgcodes below is a script that will capture screenshots with a numeric count indicator per your original post. (tested it on Ubuntu 20.04)
Script contents:
#!/bin/bash
set -uo pipefail
# add source file and line number to xtrace output
# i.e. when running: bash -x ./your_script_name.sh
export PS4='+(${BASH_SOURCE}:${LINENO}): ${FUNCNAME[0]:+${FUNCNAME[0]}(): }'
capture_screenshot() {
local output_dir="${1:-/tmp/screenshot}"
local img_name="${2:-Screenshot}"
local img_ext="${3:-png}"
# create output directory (if not already existing)
mkdir -p "${output_dir}"
# get the last image in the sorted ls output
local latest_png=$(tail -n 1 \
<(sort -n -t _ -k 2 \
<(ls ${output_dir}/*.${img_ext} 2> /dev/null)))
# use the latest image to determine img_cnt value for next image
local img_cnt=0
if [[ ${latest_png} =~ _([0-9]+)\.png ]]; then
img_cnt=$((1+${BASH_REMATCH[1]}))
elif [[ ${latest_png} =~ ${img_name}.${img_ext} ]] ; then
img_cnt=1
fi
# build path to output image
local img_path="${output_dir}/${img_name}_${img_cnt}.${img_ext}"
# remove count from output image path if count == 0
if [[ "${img_cnt}" -eq "0" ]] ; then
img_path="${output_dir}/${img_name}.${img_ext}"
fi
xfce4-screenshooter -f -s "${img_path}"
}
capture_screenshot "$#"
The uses the following as defaults, but you can change them to meet your requirements.
output directory for screenshots:
/tmp/screenshot
base screenshot image name:
Screenshot
screenshot file extension:
.png
The script will attempt to create the output directory if it does not already exist (subject to user permission for creation). Below is a sample usage.
Prior to initial script execution, the output directory does not exist:
$ ls screenshot
$
Initial execution (directory is created and Screenshot.png created:
$ ./script.sh
$ ls /tmp/screenshot/
Screenshot.png
Subsequent executions:
$ ./script.sh
$ ls /tmp/screenshot/
Screenshot_1.png Screenshot.png
$ ./script.sh
$ ls /tmp/screenshot/
Screenshot_1.png Screenshot_2.png Screenshot.png

Indeed, as suggested by #j_b in the comments, you should definitely give a try to using a timestamp with the command date +"$format".
FTR, the same idea is implemented here in this project of a gnome-screenshot bash wrapper
(disclaimer: I am the author of this repo).
Example command:
date "+%Y-%m-%d_%H-%M-%S"
↓
2021-07-29_19-13-30
So the overall script could just be something like:
#!/usr/bin/env bash
xfce4-screenshooter -f -s "$HOME/Pictures/Screenshot_$(date "+%Y-%m-%d_%H-%M-%S").png"
(Note that I added missing double-quotes, and modified your shebang, as /usr/bin/env bash is more portable than /bin/bash or /usr/bin/bash.)

Related

how to move a file after grep command when there is no return result

I wanna move a file after the grep command but as I execute my script, I noticed that there are no results coming back. regardless of that, I want to move the file/s to another directory.
this is what I've been doing:
for file in *.sup
do
grep -iq "$file" '' /desktop/list/varlogs.txt || mv "$file" /desktop/first;
done
but I am getting this error:
mv: 0653-401 Cannot rename first /desktop/first/first
suggestions would be very helpful
I am not sure what the two single quotes are for in between ..."$file" '' /desktop.... With them there, grep is looking also for $file in a file called '', so grep will throw the grep: : No such file or directory error with that there.
Also pay attention to the behavior change of adding the -q or --quiet flags, as it affects the returned value of grep and will impact whether the command to the || is run or not (see man grep for more).
I can't make out exactly what you are trying to do, but you can add a couple statements to help figure out what is going on. You could run your script with bash -x ./myscript.sh to display everything that runs as it runs, or add set -x before and set +x after the for loop in the script to show what is happening.
I added some debugging to your script and changed th || to an if/then statement to expose what is happening. Try this and see if you can find where things are going awry.
echo -e "============\nBEFORE:\n============"
echo -e "\n## The files in current dir '$(pwd)' are: ##\n$(ls)"
echo -e "\n## The files in '/desktop/first' are: ##\n$(ls /desktop/first)"
echo -e "\n## Looking for '.sup' files in '$(pwd)' ##"
for file in *.sup; do
echo -e "\n## == look for '${file}' in '/desktop/list/varlogs.txt' == ##"
# let's change this to an if/else
# the || means try the left command for success, or try the right one
# grep -iq "$file" '' /desktop/list/varlogs.txt || mv -v "$file" /desktop/first
# based on `man grep`: EXIT STATUS
# Normally the exit status is 0 if a line is selected,
# 1 if no lines were selected, and 2 if an error occurred.
# However, if the -q or --quiet or --silent is used and a line
# is selected, the exit status is 0 even if an error occurred.
# note that --ignore-case and --quiet are long versions of -i and -q/ -iq
if grep --ignore-case --quiet "${file}" '' /desktop/list/varlogs.txt; then
echo -e "\n'${file}' found in '/desktop/list/varlogs.txt'"
else
echo -e "\n'${file}' not found in '/desktop/list/varlogs.txt'"
echo -e "\nmove '${file}' to '/desktop/first'"
mv --verbose "${file}" /desktop/first
fi
done
echo -e "\n============\nAFTER:\n============"
echo -e "\n## The files in current dir '$(pwd)' are: ##\n$(ls)"
echo -e "\n## The files in '/desktop/first' are: ##\n$(ls /desktop/first)"
|| means try the first command, and if it is not successful (i.e. does not return 0), then do the next command. In your case, it appears you are looking in /desktop/list/varlogs.txt to see if any .sup files in the current directory match any in the varlogs file and if not, then move them to the /desktop/first/ directory. If matches were found, leave them in the current dir. (according to the logic you have currently)
mv --verbose explain what is being done
echo -e enables interpretation of backslash escapes
set -x shows the commands that are being run/ debugging
Please respond and clarify if anything is different. I am trying to raise in the ranks to be more helpful so I would appreciate comments, and upvotes if this was helpful.
Suggesting to avoid repeated scans of /desktop/list/varlogs.txt, and remove duplicats:
mv $(grep -o -f <<<$(ls -1 *.sup) /desktop/list/varlogs.txt|sort|uniq) /desktop/first
Suggesting to test step 1. in explanation below to list the files to be moved.
Explanation
1. grep -o -f <<<$(ls -1 *.sup) /desktop/list/varlogs.txt| sort| uniq
List all the files selected in ls -1 *.sup mentioned in /desktop/list/varlogs.txt in a single scan.
-o list only matched filenames.
<<<$(ls -1 *.sup) prepare a temporary redirected input file containing all the pattern match strings. From the output of ls -1 *.sup
|sort|uniq Than, sort the list and remove duplicates (we can move the file only once).
2. mv <files-list-output-from-step-1> /desktop/first
Move all the files found in step 1 to directory /desktop/first

Why is a part of the code inside a (False) if statement executed?

I wrote a small script which:
prints the content of a file (generated by another application) on paper with a matrix printer
prints the same line into a backup file
removes the original file.
The script runs every minute by a cronjob and works fine as long as there are files to print. If there are no files to print, it prints an empty line on the matrix printer and in the backup file. I don't understand why this happens as i implemented an if statement which checks if there is a file to print before the print command is executed. This behaviour only happens if the script is executed by the cron and not if i execute it manually with ./script.sh. What's the reason of this? and how can i solve it?
Something i noticed on the side is that if I place an echo "hi" command in the script, its printed to the matrix printer and the backup file. I expected that its printed to the console console when it has no >> something behind. How does this work?
The script:
#!/bin/bash
# Make sure the backup directory exists
if [ ! -d /home/user/backup_logprint ]
then
mkdir /home/user/backup_logprint
fi
# Print the records if there are any
date=`date +%Y-%m-%d`
filename='_logprint_backup'
printer_path="/dev/usb/lp0"
if [ `ls /tmp/ | grep logprint | wc -l` -gt 0 ]
then
for f in `ls /tmp | grep logprint`
do
echo `cat /tmp/$f` >> "/home/user/backup_logprint/$date$filename"
echo `cat /tmp/$f` >> $printer_path
rm "/tmp/$f"
done
fi
There's no need for ls or an if statement. Just use a proper glob in the for loop, and if no file match, the loop won't be entered.
#!/bin/bash
# Don't check first; just let mkdir decide if
# anything actually needs to be created.
d=/home/user/backup_logprint
mkdir -p "$d"
filename=$(date +"$d/%Y-%m-%d_logprint_backup")
printer_path="/dev/usb/lp0"
# Cause non-matching globs to expand to an empty
# sequence instead of being treated literally.
shopt -s nullglob
for f in /tmp/*logprint*; do
cat "$f" > "$printer_path" && mv "$f" "$d"
done

List files greater than 100K in bash

I want to list the files recursively in the HOME directory. I'm trying to write my own script , so I should not use the command find or ls. My script is:
#!/bin/bash
minSize=102400;
printFiles() {
for x in "$1/"*; do
if [ -d "$x" ]; then
printFiles "$x";
else
size=$(wc -c "$x");
if [[ "$size" -gt "$minSize" ]]; then
echo "$size";
fi
fi
done
}
printFiles "/~";
So, the problem here is that when I run this script, the terminal throws Line 11: division by 0 and /home/gandalf/Videos/*: No such file or directory. I have not divided by any number, why I'm getting this error?. And the second one?
Alternatively, I can't use find or ls because I have to display the files one by one asking to the user if he want to see the next file or not. This is possible using the command find or ls or only can be done writing my own function?
Thanks.
size=$(wc -c "$x");
That's the line that is failing. When you run that wc command manually you should be able to see why:
$ wc -c /tmp/out
5 /tmp/out
The output contains not only the file size but also the file name. So you can't use $size with the -gt comparator on the next line. One way to fix that is to change the wc line to use cut (or awk, or sed, etc) to keep just the file size.
size=$(wc -c "$x" | cut -f1 -d " ")
A simpler alternative suggested by #mklement0:
size=$(wc -c < "$x")

Bash Variable Maths Not Working

I have a simple bash script, which forms part of an in house web app that I've developed.
It's purpose is to automate deletion of thumbnails of images when the original image has been deleted by the user.
The script logs some basic status info to a file /var/log/images.log
#!/bin/bash
cd $thumbpath
filecount=0
# Purge extraneous thumbs
find . -type f | while read file
do
if [ ! -f "$imagepath/$file" ]
then
filecount=$[$filecount+1]
rm -f "$file"
fi
done
echo `date`: $filecount extraneous thumbs removed>>/var/log/images.log
Whilst the script correctly deletes thumbs, it doesn't correctly output the number of thumbs that are being purged, it always shows 0.
For example, having just manually created some orphaned thumbnails, and then running my script, the manually generated orphaned thumbs are deleted, but the log shows:
Thu Jun 9 23:30:12 BST 2011: 0 extraneous thumbs removed
What am I doing wrong that is stopping $filecounter from showing a number other than zero, when files are being deleted.
I've created the following bash script to test this, and this works perfectly, outputting 0 then 1:
#!/bin/bash
count=0
echo $count
count=$[$count+1]
echo $count
Edit:
Thanks for the answers, but why does the following work
$ x=3
$ x=$[$x+1]
$ echo $x
4
...and also the second example works, yet it doesn't work in the first script?
Second Edit:
This works
count=0
echo Initial Value $count
for i in `seq 1 5`
do
count=$[$count+1]
echo $count
done
echo Final Value $count
Initial Value 0
1
2
3
4
5
Final Value 5
as does replacing count=$[$count+1] with count=$((count+1)), but not in my initial script.
You're using the wrong operator. Try using $(( ... )) instead, e.g.:
$ x=4
$ y=$((x + 1))
$ echo $y
5
$
EDIT
The other problem you're bumping into is down to the pipe. Bumped into this one before (with ksh, but wouldn't suprise me to find that other shells have the same problem). The pipe is forking another bash process, so when you do the increment, filcount is getting incremented in the subshell that's been forked after the pipe. This value isn't passed back to the calling shell as the subshell has it's own independent environment (environment variables are inherited in called processes, but called process cannot modify the environment of the calling process).
As an example, this demonstrates that filecount gets incremented okay:
#!/bin/bash
filecount=0
ls /bin | while read x
do
filecount=$((filecount + 1))
echo $filecount
done
echo $filecount
...so you should see filecount increase in the loop, but the final filecount will be zero because this echo belongs to the main shell, but the forked subshell (which consists purely of the while loop).
One way you can get the value back is like this...
#!/bin/bash
filecount=0
filecount=`ls /bin | while read x
do
filecount=$((filecount + 1))
echo $filecount
done | tail -1`
echo $filecount
This will only work if you don't care about any other stdout output in the loop as this throws it all away apart from the last line we output (the final value of filecount). This works because we're using stdout and stdin to feed the data back to the parent shell.
Depending on your viewpoint this is either a nasty hack or a nifty bit of shell jiggery-pokery. I'll leave you to decide what you think it is :-)
If you remove the pipeline into the while construct, you remove bash's need to create a subshell.
Change this:
filecount=0
find . -type f | while read file; do
if [ ! -f "$imagepath/$file" ]; then
filecount=$[$filecount+1]
rm -f "$file"
fi
done
echo $filecount
to this:
filecount=0
while read file; do
if [ ! -f "$imagepath/$file" ]; then
rm -f "$file" && (( filecount++ ))
fi
done < <(find . -type f)
echo $filecount
That is harder to read because the find command is hidden at the end. Another possibility is:
files=$( find . -type f )
while ...; do
:
done <<< "$files"
Chris J is quite right that you are using the wrong operator and POSIX subshell variable scoping means you can't get a final count that way.
As a side note, when doing math operations you could also consider using the let shell bultin like this:
$ filecount=4
$ let filecount=$filecount+1
$ echo $filecount
5
Also if you want scoping to just work like you expected it to in spite of that pipeline, you could use zsh instead of bash. In this case it should be a drop in replacement and work as expected.

Equivalent of %~dp0 (retrieving source file name) in sh

I'm converting some Windows batch files to Unix scripts using sh. I have problems because some behavior is dependent on the %~dp0 macro available in batch files.
Is there any sh equivalent to this? Any way to obtain the directory where the executing script lives?
The problem (for you) with $0 is that it is set to whatever command line was use to invoke the script, not the location of the script itself. This can make it difficult to get the full path of the directory containing the script which is what you get from %~dp0 in a Windows batch file.
For example, consider the following script, dollar.sh:
#!/bin/bash
echo $0
If you'd run it you'll get the following output:
# ./dollar.sh
./dollar.sh
# /tmp/dollar.sh
/tmp/dollar.sh
So to get the fully qualified directory name of a script I do the following:
cd `dirname $0`
SCRIPTDIR=`pwd`
cd -
This works as follows:
cd to the directory of the script, using either the relative or absolute path from the command line.
Gets the absolute path of this directory and stores it in SCRIPTDIR.
Goes back to the previous working directory using "cd -".
Yes, you can! It's in the arguments. :)
look at
${0}
combining that with
{$var%Pattern}
Remove from $var the shortest part of $Pattern that matches the back end of $var.
what you want is just
${0%/*}
I recommend the Advanced Bash Scripting Guide
(that is also where the above information is from).
Especiall the part on Converting DOS Batch Files to Shell Scripts
might be useful for you. :)
If I have misunderstood you, you may have to combine that with the output of "pwd". Since it only contains the path the script was called with!
Try the following script:
#!/bin/bash
called_path=${0%/*}
stripped=${called_path#[^/]*}
real_path=`pwd`$stripped
echo "called path: $called_path"
echo "stripped: $stripped"
echo "pwd: `pwd`"
echo "real path: $real_path
This needs some work though.
I recommend using Dave Webb's approach unless that is impossible.
In bash under linux you can get the full path to the command with:
readlink /proc/$$/fd/255
and to get the directory:
dir=$(dirname $(readlink /proc/$$/fd/255))
It's ugly, but I have yet to find another way.
I was trying to find the path for a script that was being sourced from another script. And that was my problem, when sourcing the text just gets copied into the calling script, so $0 always returns information about the calling script.
I found a workaround, that only works in bash, $BASH_SOURCE always has the info about the script in which it is referred to. Even if the script is sourced it is correctly resolved to the original (sourced) script.
The correct answer is this one:
How do I determine the location of my script? I want to read some config files from the same place.
It is important to realize that in the general case, this problem has no solution. Any approach you might have heard of, and any approach that will be detailed below, has flaws and will only work in specific cases. First and foremost, try to avoid the problem entirely by not depending on the location of your script!
Before we dive into solutions, let's clear up some misunderstandings. It is important to understand that:
Your script does not actually have a location! Wherever the bytes end up coming from, there is no "one canonical path" for it. Never.
$0 is NOT the answer to your problem. If you think it is, you can either stop reading and write more bugs, or you can accept this and read on.
...
Try this:
${0%/*}
This should work for bash shell:
dir=$(dirname $(readlink -m $BASH_SOURCE))
Test script:
#!/bin/bash
echo $(dirname $(readlink -m $BASH_SOURCE))
Run test:
$ ./somedir/test.sh
/tmp/somedir
$ source ./somedir/test.sh
/tmp/somedir
$ bash ./somedir/test.sh
/tmp/somedir
$ . ./somedir/test.sh
/tmp/somedir
This is a script can get the shell file real path when executed or sourced.
Tested in bash, zsh, ksh, dash.
BTW: you shall clean the verbose code by yourself.
#!/usr/bin/env bash
echo "---------------- GET SELF PATH ----------------"
echo "NOW \$(pwd) >>> $(pwd)"
ORIGINAL_PWD_GETSELFPATHVAR=$(pwd)
echo "NOW \$0 >>> $0"
echo "NOW \$_ >>> $_"
echo "NOW \${0##*/} >>> ${0##*/}"
if test -n "$BASH"; then
echo "RUNNING IN BASH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${BASH_SOURCE[0]}
elif test -n "$ZSH_NAME"; then
echo "RUNNING IN ZSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${(%):-%x}
elif test -n "$KSH_VERSION"; then
echo "RUNNING IN KSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${.sh.file}
else
echo "RUNNING IN DASH OR OTHERS ELSE..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=$(lsof -p $$ -Fn0 | tr -d '\0' | grep "${0##*/}" | tail -1 | sed 's/^[^\/]*//g')
fi
echo "EXECUTING FILE PATH: $SH_FILE_RUN_PATH_GETSELFPATHVAR"
cd "$(dirname "$SH_FILE_RUN_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_RUN_PATH_GETSELFPATHVAR")
# Iterate down a (possible) chain of symlinks as lsof of macOS doesn't have -f option.
while [ -L "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR" ]; do
SH_FILE_REAL_PATH_GETSELFPATHVAR=$(readlink "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR")
cd "$(dirname "$SH_FILE_REAL_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_REAL_PATH_GETSELFPATHVAR")
done
# Compute the canonicalized name by finding the physical path
# for the directory we're in and appending the target file.
SH_SELF_PATH_DIR_RESULT=$(pwd -P)
SH_FILE_REAL_PATH_GETSELFPATHVAR=$SH_SELF_PATH_DIR_RESULT/$SH_FILE_RUN_BASENAME_GETSELFPATHVAR
echo "EXECUTING REAL PATH: $SH_FILE_REAL_PATH_GETSELFPATHVAR"
echo "EXECUTING FILE DIR: $SH_SELF_PATH_DIR_RESULT"
cd "$ORIGINAL_PWD_GETSELFPATHVAR" || return 1
unset ORIGINAL_PWD_GETSELFPATHVAR
unset SH_FILE_RUN_PATH_GETSELFPATHVAR
unset SH_FILE_RUN_BASENAME_GETSELFPATHVAR
unset SH_FILE_REAL_PATH_GETSELFPATHVAR
echo "---------------- GET SELF PATH ----------------"
# USE $SH_SELF_PATH_DIR_RESULT BEBLOW
I have tried $0 before, namely:
dirname $0
and it just returns "." even when the script is being sourced by another script:
. ../somedir/somescript.sh

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