I am trying to make a script what looks at a folder and will automatically encode files that go into that folder using hand brake. I want to do this doing monitoring the folder using inotify putting the new additions to the folder into a list then using a cron job to encode them overnight. However when using a while loop to loop over the list handbrake only encodes the first file exists then the scripts carrys on to after the loop without doing every file in the list. Here is the script that is calling handbrake:
#!/bin/bash
while IFS= read -r line
do
echo "$(basename "$line")"
HandBrakeCLI -Z "Very Fast 1080p30" -i "$line" -o "$line.m4v"
rm "$line"
done < list.txt
> list.txt
When testing the loop with a simple echo instead of the HandBrakeCLI it works fine and prints out every file so I have no idea what is wrong.
Here is the scripts that is monitoring the folder incase that is the problem:
#!/bin/bash
if ! [ -f list.txt ]
then
touch list.txt
fi
inotifywait -m -e create --format "%w%f" tv-shows | while read FILE
do
echo "$FILE" >> list.txt
done
Any help would be great, thanks
EDIT:
Just to be more specific, the script works fine for the first file in the list.txt, it encodes it no problem and removes the old version, but then it doesn't do any of the others in the list
Taken from here
To solve the problem simply
echo "" | HandBrakeCLI ......
or
HandBrakeCLI ...... < /dev/null
Related
So guys,
I need your help trying to identify the fastest and the most "fault" tolerant solution to my problem.
I have a shell script which executes some functions, based on a txt file, in which I have a list of files.
The list can contain from 1 file to X files.
What I would like to do is iterate over the content of the file and execute my scripts for only 4 items out of the file.
Once the functions have been executed for these 4 files, go over to the next 4 .... and keep on doing so until all the files from the list have been "processed".
My code so far is as follows.
#!/bin/bash
number_of_files_in_folder=$(cat list.txt | wc -l)
max_number_of_files_to_process=4
Translated_files=/home/german_translated_files/
while IFS= read -r files
do
while [[ $number_of_files_in_folder -gt 0 ]]; do
i=1
while [[ $i -le $max_number_of_files_to_process ]]; do
my_first_function "$files" & # I execute my translation function for each file, as it can only perform 1 file per execution
find /home/german_translator/ -name '*.logs' -exec mv {} $Translated_files \; # As there will be several files generated, I have them copied to another folder
sed -i "/$files/d" list.txt # We remove the processed file from within our list.txt file.
my_second_function # Without parameters as it will process all the files copied at step 2.
done
# here, I want to have all the files processed and don't stop after the first iteration
done
done < list.txt
Unfortunately, as I am not quite good at shell scripting, I do not know how to structure it so that it won't waste any resources and mostly, to make sure that it "processes" everything from that file.
Do you have any advice on how to achieve what I am trying to achieve?
only 4 items out of the file. Once the functions have been executed for these 4 files, go over to the next 4
Seems to be quite easy with xargs.
your_function() {
echo "Do something with $1 $2 $3 $4"
}
export -f your_function
xargs -d '\n' -n 4 bash -c 'your_function "$#"' _ < list.txt
xargs -d '\n' for each line
-n 4 take for arguments
bash .... - run this command with 4 arguments
_ - the syntax is bash -c <script> $0 $1 $2 etc..., see man bash.
"$#" - forward arguments
export -f your_function - export your function to environment so child bash can pick it up.
I execute my translation function for each file
So you execute your translation function for each file, not for each 4 files. If the "translation function" is really for each file with no inter-file state, consider rather executing 4 processes in parallel with same code and just xargs -P 4.
If you have GNU Parallel it looks something like this:
doit() {
my_first_function "$1"
my_first_function "$2"
my_first_function "$3"
my_first_function "$4"
my_second_function "$1" "$2" "$3" "$4"
}
export -f doit
cat list.txt | parallel -n4 doit
I wrote a small script which:
prints the content of a file (generated by another application) on paper with a matrix printer
prints the same line into a backup file
removes the original file.
The script runs every minute by a cronjob and works fine as long as there are files to print. If there are no files to print, it prints an empty line on the matrix printer and in the backup file. I don't understand why this happens as i implemented an if statement which checks if there is a file to print before the print command is executed. This behaviour only happens if the script is executed by the cron and not if i execute it manually with ./script.sh. What's the reason of this? and how can i solve it?
Something i noticed on the side is that if I place an echo "hi" command in the script, its printed to the matrix printer and the backup file. I expected that its printed to the console console when it has no >> something behind. How does this work?
The script:
#!/bin/bash
# Make sure the backup directory exists
if [ ! -d /home/user/backup_logprint ]
then
mkdir /home/user/backup_logprint
fi
# Print the records if there are any
date=`date +%Y-%m-%d`
filename='_logprint_backup'
printer_path="/dev/usb/lp0"
if [ `ls /tmp/ | grep logprint | wc -l` -gt 0 ]
then
for f in `ls /tmp | grep logprint`
do
echo `cat /tmp/$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
I am looking to create a bash script that keeps checking a file in directory and perform certain operation on it. I am using while loop, if file does not exists I want that while loop stays quite and keeps on checking condition. Here is what i created but it keeps throwing error that file not found, if file is not there.
while [ ! -f /home/master/applications/tmp/mydata.txt ]
do
cat mydata.txt;
rm mydata.txt;
sleep 1; done
There are two issue in your implementation:
You should use the same (absolute or relative) path in your while loop test statement [ ! -f $file ] and in your cat and rm commands.
The cat command is looking for the file in the current working directory (pwd) and your while statement might be looking somewhere else and hence, your implementation is buggy and won't work as expected if your pwd isn't /home/master/applications/tmp.
You need to move your cat command and rm command after the while block. It doesn't make sense to cat a file if a file doesn't exist. I think your misplaced those commands.
Try this:
file="/home/master/applications/tmp/mydata.txt"
while [ ! -f "$file" ]
do
sleep 1
done
cat $file
rm $file
EDIT
As per suggestion from #Ivan, you could use until instead of while as it suits more to your requirements.
file="/home/master/applications/tmp/mydata.txt"
until [ -f "$file" ]; do sleep 1; done
cat $file
rm $file
Making a different assumption than abhiarora, I'll guess maybe you meant for the file to reappear, and you want it shown each time.
file=/home/master/applications/tmp/mydata.txt
while :
do if [[ -f "$file" ]]
then echo "$(<"$file")"
rm "$file"
fi
sleep 1
done
This creates an infinite loop. If that's NOT what you wanted, use abhiarora's solution.
I have multiple files with an insanely long list of commands. I can't run them all in one go, so I need a smart way to read and execute from file as well as delete the command after completion.
So far I have tried
for i in filename.txt ; do ; execute $i ; sed -s 's/$i//' ; done ;
but it doesn't work. Before I introduced sed, $i was executing. Now even that is not working.
I thought of a workaround where I will read first line and delete first line till file is empty.
Any better ideas or commands?
This should work for you, list.txt is your file containing commands.
Make sure you backup the command file before running.
while read line; do $line;sed -i '1d' list.txt;done < "list.txt"
sed -i edits in-place so list.txt will be changed along the loop and you will end up with a empty file.
I think what you want to do is something like this:
while read -r -- i; do $i; sed -i "0,/$i/s/$i//;/^$/d" filename.txt; done < filename.txt
The file is read into the loop. Each line is executed, and the sed command will delete only the first entry it finds, then delete the empty line.
I think that one way to do it is to have the source file of all the commands to be executed, and the script that executes the commands also writes a second log file that lists the files as they are executed.
If you need to resume the process, you work on the lines in the source file that are not present in the log file.
logfile=commands.log
srcfile=commands.src
oldfile=commands.old
trap "mv $oldfile $logfile; exit 1" 0 1 2 3 13 15
[ -f $logfile ] || cp /dev/null $logfile
cp $logfile $oldfile
comm -23 $srcfile $logfile |
while read -r line
do
echo "$line" >> $oldfile
($line) < /dev/null
done
mv $oldfile $logfile
trap 0
I am running this command in a script
while [ 1 ]
do
if [ -e $LOG ]
then
grep -A 5 -B 5 -f $PATTERNS $LOG >> $FOREMAIL
break
fi
done
$LOG file is scp'ed from another machine. So as soon as it appears in the current directory, while loop detects it and does the grep. The problem is, the $FOREMAIL file turns up to be empty. But if I run this grep outside of the script as a standalone command with same files and params, I can see that it generates an output.
I am baffled as to why this command is generating no o/p in the script?
The -e is triggering as soon as scp creates the file, while it still has no data in it, and grep is operating on an empty file. You need to wait until the file has finished transferring.
You could accomplish this by transferring to a temporary filename, than running mv over ssh from the machine which is pushing the file up.
Edit: the code for the machine copying to log file up...
scp $log 192.168.0.1:/logfiles/${log}.tmp
ssh 192.168.0.1 mv /logfiles/${log}.tmp /logfiles/${log}
Before you can grep, you need to wait for two things: 1) the download started (file comes into existence) and 2) download finished (nobody is opening the file anymore). I have a script call waitfor.sh, which does this:
#!/bin/bash
# waitfor.sh - wait for a file fully downloaded (via Firefox, scp, ...)
# Syntax:
# waitfor.sh filename
FILENAME=$1 # Name of file to wait for
INTERVAL=10 # Wait interval of N seconds
# Wait for download started
while [ ! -f $FILENAME ]
do
sleep $INTERVAL
done
# Wait for download finished
while lsof $FILENAME
do
sleep $INTERVAL
done
To use it:
waitfor.sh $LOG
grep ...
Could it be that the while [1] is very fast, so when the file starts copying, it shows up as an empty file first before copying is complete? Depending on the size of the file, try a sleep delay inside the then loop. Figuring out when a file finishes copying when done by an external process is probably a separate question - e.g. googling for something like "how to tell when scp has finshed copying a file" turns up a bunch of links like: https://superuser.com/questions/45224/is-there-a-way-to-tell-if-a-file-is-done-copying
Better to use:
if [ -f $LOG ]
instead of:
if [ -e $LOG ]
-f checks for a regular type
-e checks for any file
Here's what I ended up doing:
scp $LOGFILE
then
scp $SCPDONE # empty file
And modified the if clause like this:
while [ 1 ]
do
if [ -e $SCPDONE ]
then
grep -A 5 -B 5 -f $PATTERNS $LOG >> $FOREMAIL
break
fi
done