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 "$file"
fi
sleep 1
done
This creates an infinite loop. If that's NOT what you wanted, use abhiarora's solution.
Related
I wrote a small script which:
prints the content of a file (generated by another application) on paper with a matrix printer
prints the same line into a backup file
removes the original file.
The script runs every minute by a cronjob and works fine as long as there are files to print. If there are no files to print, it prints an empty line on the matrix printer and in the backup file. I don't understand why this happens as i implemented an if statement which checks if there is a file to print before the print command is executed. This behaviour only happens if the script is executed by the cron and not if i execute it manually with ./script.sh. What's the reason of this? and how can i solve it?
Something i noticed on the side is that if I place an echo "hi" command in the script, its printed to the matrix printer and the backup file. I expected that its printed to the console console when it has no >> something behind. How does this work?
The script:
#!/bin/bash
# Make sure the backup directory exists
if [ ! -d /home/user/backup_logprint ]
then
mkdir /home/user/backup_logprint
fi
# Print the records if there are any
date=`date +%Y-%m-%d`
filename='_logprint_backup'
printer_path="/dev/usb/lp0"
if [ `ls /tmp/ | grep logprint | wc -l` -gt 0 ]
then
for f in `ls /tmp | grep logprint`
do
echo `cat /tmp/$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" "$d"
done
I am trying to make a script what looks at a folder and will automatically encode files that go into that folder using hand brake. I want to do this doing monitoring the folder using inotify putting the new additions to the folder into a list then using a cron job to encode them overnight. However when using a while loop to loop over the list handbrake only encodes the first file exists then the scripts carrys on to after the loop without doing every file in the list. Here is the script that is calling handbrake:
#!/bin/bash
while IFS= read -r line
do
echo "$(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
I have people uploading files to a directory on my Ubuntu Server.
I need to move those files to the final location (another directory) only when I know these files are fully uploaded.
Here's my script so far:
#!/bin/bash
cd /var/uploaded_by_users
for filename in *; do
lsof $filename
if [ -z $? ]; then
# file has been closed, move it
else
echo "*** File is open. Skipping..."
fi
done
cd -
However it's not working as it says some files are open when that's not true. I supposed $? would have 0 if the file was closed and 1 if it wasn't but I think that's wrong.
I'm not linux expert so I'm looking to know how to implement this simple script that will run on a cron job every 1 minute.
[ -z $? ] checks if $? is of zero length or not. Since $? will never be a null string, your check will always fail and result in else part being executed.
You need to test for numeric zero, as below:
lsof "$filename" >/dev/null; lsof_status=$?
if [ "$lsof_status" -eq 0 ]; then
# file is open, skipping
else
# move it
fi
Or more simply (as Benjamin pointed out):
if lsof "$filename" >/dev/null; then
# file is open, skip
else
# move it
fi
Using negation, we can shorten the if statement (as dimo414 pointed out):
if ! lsof "$filename" >/dev/null; then
# move it
fi
You can shorten it even further, using &&:
for filename in *; do
lsof "$filename" >/dev/null && continue # skip if the file is open
# move the file
done
You may not need to worry about when the write is complete, if you are moving the file to a different location in the same file system. As long as the client is using the same file descriptor to write to the file, you can simply create a new hard link for the upload file, then remove the original link. The client's file descriptor won't be affected by one of the links being removed.
cd /var/uploaded_by_users
for f in *; do
ln "$f" /somewhere/else/"$f"
rm "$f"
done
I have a bash program that will write to an output file. This file may or may not exist, but the script must check permissions and fail early. I can't find an elegant way to make this happen. Here's what I have tried.
set +e
touch $file
set -e
if [ $? -ne 0 ]; then exit;fi
I keep set -e on for this script so it fails if there is ever an error on any line. Is there an easier way to do the above script?
Why complicate things?
file=exists_and_writeable
if [ ! -e "$file" ] ; then
touch "$file"
fi
if [ ! -w "$file" ] ; then
echo cannot write to $file
exit 1
fi
Or, more concisely,
( [ -e "$file" ] || touch "$file" ) && [ ! -w "$file" ] && echo cannot write to $file && exit 1
Rather than check $? on a different line, check the return value immediately like this:
touch file || exit
As long as your umask doesn't restrict the write bit from being set, you can just rely on the return value of touch
You can use -w to check if a file is writable (search for it in the bash man page).
if [[ ! -w $file ]]; then exit; fi
Why must the script fail early? By separating the writable test and the file open() you introduce a race condition. Instead, why not try to open (truncate/append) the file for writing, and deal with the error if it occurs? Something like:
$ echo foo > output.txt
$ if [ $? -ne 0 ]; then die("Couldn't echo foo")
As others mention, the "noclobber" option might be useful if you want to avoid overwriting existing files.
Open the file for writing. In the shell, this is done with an output redirection. You can redirect the shell's standard output by putting the redirection on the exec built-in with no argument.
set -e
exec >shell.out # exit if shell.out can't be opened
echo "This will appear in shell.out"
Make sure you haven't set the noclobber option (which is useful interactively but often unusable in scripts). Use > if you want to truncate the file if it exists, and >> if you want to append instead.
If you only want to test permissions, you can run : >foo.out to create the file (or truncate it if it exists).
If you only want some commands to write to the file, open it on some other descriptor, then redirect as needed.
set -e
exec 3>foo.out
echo "This will appear on the standard output"
echo >&3 "This will appear in foo.out"
echo "This will appear both on standard output and in foo.out" | tee /dev/fd/3
(/dev/fd is not supported everywhere; it's available at least on Linux, *BSD, Solaris and Cygwin.)
I originally had a set of images of the form image_001.jpg, image_002.jpg, ...
I went through them and removed several. Now I'd like to rename the leftover files back to image_001.jpg, image_002.jpg, ...
Is there a Linux command that will do this neatly? I'm familiar with rename but can't see anything to order file names like this. I'm thinking that since ls *.jpg lists the files in order (with gaps), the solution would be to pass the output of that into a bash loop or something?
If I understand right, you have e.g. image_001.jpg, image_003.jpg, image_005.jpg, and you want to rename to image_001.jpg, image_002.jpg, image_003.jpg.
EDIT: This is modified to put the temp file in the current directory. As Stephan202 noted, this can make a significant difference if temp is on a different filesystem. To avoid hitting the temp file in the loop, it now goes through image*
i=1; temp=$(mktemp -p .); for file in image*
do
mv "$file" $temp;
mv $temp $(printf "image_%0.3d.jpg" $i)
i=$((i + 1))
done
A simple loop (test with echo, execute with mv):
I=1
for F in *; do
echo "$F" `printf image_%03d.jpg $I`
#mv "$F" `printf image_%03d.jpg $I` 2>/dev/null || true
I=$((I + 1))
done
(I added 2>/dev/null || true to suppress warnings about identical source and target files. If this is not to your liking, go with Matthew Flaschen's answer.)
Some good answers here already; but some rely on hiding errors which is not a good idea (that assumes mv will only error because of a condition that is expected - what about all the other reaons mv might error?).
Moreover, it can be done a little shorter and should be better quoted:
for file in *; do
printf -vsequenceImage 'image_%03d.jpg' "$((++i))"
[[ -e $sequenceImage ]] || \
mv "$file" "$sequenceImage"
done
Also note that you shouldn't capitalize your variables in bash scripts.
Try the following script:
numerate.sh
This code snipped should do the job:
./numerate.sh -d <your image folder> -b <start number> -L 3 -p image_ -s .jpg -o numerically -r
This does the reverse of what you are asking (taking files of the form *.jpg.001 and converting them to *.001.jpg), but can easily be modified for your purpose:
for file in *
do
if [[ "$file" =~ "(.*)\.([[:alpha:]]+)\.([[:digit:]]{3,})$" ]]
then
mv "${BASH_REMATCH[0]}" "${BASH_REMATCH[1]}.${BASH_REMATCH[3]}.${BASH_REMATCH[2]}"
fi
done
I was going to suggest something like the above using a for loop, an iterator, cut -f1 -d "_", then mv i i.iterator. It looks like it's already covered other ways, though.