basename and convert in bash script error - linux

I am a beginner in bash script programming. I'm trying to make a function to convert a large number of photo files without changing the name. I have read several topics about using basename or {s%.*} to remove the type, but without succeeding in applying it.
thanks in advance for your help
#!/bin/bash
FOLDER=~/Bureau/data/jpg
ERROR="$FOLDER/error.txt"
CURRENT=$PWD
if ! [[ -d "$FOLDER" ]]|| ! [[ -e $ERROR ]]
then
mkdir -p $FOLDER
touch $ERROR
else
rm -f "$FOLDER/*"
touch $ERROR
fi
for img in "$CURRENT/*.tif"; do
filename=$(basename -- "$img")
filename="${filename%.*}"
convert $img "$FOLDER/$filename.jpg" 2> $ERROR
done
Error images

I think you can simplify.
#!/bin/bash
dir="~/Bureau/data/jpg"
rm -fr "$dir/" # -f doesn't fail if it didn't exist
mkdir -p "$dir" # -p doesn't fail if it *did* exist
for img in "$PWD"/*.tif; do
stub="${img%.tif}" # simple strip the end
convert "$img" "$dir/$stub.jpg" 2> "$stub.err # err log per file
done # 2> "$dir/err.log" # if you don't want an error per, use this instead
You have a lot of cases of globs in quotes, which doesn't work.
Use "$PWD"/*.tif instead of "$PWD/*.tif".
Also, 2> inside the loop overwrites. Either use 2>>, or 2> outside the loop.

Related

How can I batch rename multiple images with their path names and reordered sequences in bash?

My pictures are kept in the folder with the picture-date for folder name, for example the original path and file names:
.../Pics/2016_11_13/wedding/DSC0215.jpg
.../Pics/2016_11_13/afterparty/DSC0234.jpg
.../Pics/2016_11_13/afterparty/DSC0322.jpg
How do I rename the pictures into the format below, with continuous sequences and 4-digit padding?
.../Pics/2016_11_13_wedding.0001.jpg
.../Pics/2016_11_13_afterparty.0002.jpg
.../Pics/2016_11_13_afterparty.0003.jpg
I'm using Bash 4.1, so only mv command is available. Here is what I have now but it's not working
#!/bin/bash
p=0
for i in *.jpg;
do
mv "$i" "$dirname.%03d$p.JPG"
((p++))
done
exit 0
Let say you have something like .../Pics/2016_11_13/wedding/XXXXXX.jpg; then go in directory .../Pics/2016_11_13; from there, you should have a bunch of subdirectories like wedding, afterparty, and so on. Launch this script (disclaimer: I didn't test it):
#!/bin/sh
for subdir in *; do # scan directory
[ ! -d "$subdir" ] && continue; # skip non-directory
prognum=0; # progressive number
for file in $(ls "$dir"); do # scan subdirectory
(( prognum=$prognum+1 )) # increment progressive
newname=$(printf %4.4d $prognum) # format it
newname="$subdir.$newname.jpg" # compose the new name
if [ -f "$newname" ]; then # check to not overwrite anything
echo "error: $newname already exist."
exit
fi
# do the job, move or copy
cp "$subdir/$file" "$newname"
done
done
Please note that I skipped the "date" (2016_11_13) part - I am not sure about it. If you have a single date, then it is easy to add these digits in # compose the new name. If you have several dates, then you can add a nested for for scanning the "date" directories. One more reason I skipped this, is to let you develop something by yourself, something you can be proud of...
Using only mv and bash builtins:
#! /bin/bash
shopt -s globstar
cd Pics
p=1
# recursive glob for .jpg files
for i in **/*.jpg
do
# (date)/(event)/(filename).jpg
if [[ $i =~ (.*)/(.*)/(.*).jpg ]]
then
newname=$(printf "%s_%s.%04d.jpg" "${BASH_REMATCH[#]:1:2}" "$p")
echo mv "$i" "$newname"
((p++))
fi
done
globstar is a bash 4.0 feature, and regex matching is available even in OSX's anitque bash.

linux zip and exclude dir via bash/shell script

I am trying to write a bash/shell script to zip up a specific folder and ignore certain sub-dirs in that folder.
This is the folder I am trying to zip "sync_test5":
My bash script generates an ignore list (based on) and calls the zip function like this:
#!/bin/bash
SYNC_WEB_ROOT_BASE_DIR="/home/www-data/public_html"
SYNC_WEB_ROOT_BACKUP_DIR="sync_test5"
SYNC_WEB_ROOT_IGNORE_DIR="dir_to_ignore dir2_to_ignore"
ignorelist=""
if [ "$SYNC_WEB_ROOT_IGNORE_DIR" != "" ];
then
for ignoredir in $SYNC_WEB_ROOT_IGNORE_DIR
do
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/**\*"
done
fi
FILE="$SYNC_BACKUP_DIR/$DATETIMENOW.website.zip"
cd $SYNC_WEB_ROOT_BASE_DIR;
zip -r $FILE $SYNC_WEB_ROOT_BACKUP_DIR -x $ignorelist >/dev/null
echo "Done"
Now this script runs without error, however it is not ignoring/excluding the dirs I've specified.
So, I had the shell script output the command it tried to run, which was:
zip -r 12-08-2014_072810.website.zip sync_test5 -x sync_test5/dir_to_ignore/**\* sync_test5/dir2_to_ignore/**\*
Now If I run the above command directly in putty like this, it works:
So, why doesn't my shell script exclude working as intended? the command that is being executed is identical (in shell and putty directly).
Because backslash quotings in a variable after word splitting are not evaluated.
If you have a='123\4', echo $a would give
123\4
But if you do it directly like echo 123\4, you'd get
1234
Clearly the arguments you pass with the variable and without the variables are different.
You probably just meant to not quote your argument with backslash:
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/***"
Btw, what actual works is a non-evaluated glob pattern:
zip -r 12-08-2014_072810.website.zip sync_test5 -x 'sync_test5/dir_to_ignore/***' 'sync_test5/dir2_to_ignore/***'
You can verify this with
echo zip -r 12-08-2014_072810.website.zip sync_test5 -x sync_test5/dir_to_ignore/**\* sync_test5/dir2_to_ignore/**\*
And this is my suggestion:
#!/bin/bash
SYNC_WEB_ROOT_BASE_DIR="/home/www-data/public_html"
SYNC_WEB_ROOT_BACKUP_DIR="sync_test5"
SYNC_WEB_ROOT_IGNORE_DIR=("dir_to_ignore" "dir2_to_ignore")
IGNORE_LIST=()
if [[ -n $SYNC_WEB_ROOT_IGNORE_DIR ]]; then
for IGNORE_DIR in "${SYNC_WEB_ROOT_IGNORE_DIR[#]}"; do
IGNORE_LIST+=("$SYNC_WEB_ROOT_BACKUP_DIR/$IGNORE_DIR/***") ## "$SYNC_WEB_ROOT_BACKUP_DIR/$IGNORE_DIR/*" perhaps is enough?
done
fi
FILE="$SYNC_BACKUP_DIR/$DATETIMENOW.website.zip" ## Where is $SYNC_BACKUP_DIR set?
cd "$SYNC_WEB_ROOT_BASE_DIR";
zip -r "$FILE" "$SYNC_WEB_ROOT_BACKUP_DIR" -x "${IGNORE_LIST[#]}" >/dev/null
echo "Done"
This is what I ended up with:
#!/bin/bash
# This script zips a directory, excluding specified files, types and subdirectories.
# while zipping the directory it excludes hidden directories and certain file types
[[ "`/usr/bin/tty`" == "not a tty" ]] && . ~/.bash_profile
DIRECTORY=$(cd `dirname $0` && pwd)
if [[ -z $1 ]]; then
echo "Usage: managed_directory_compressor /your-directory/ zip-file-name"
else
DIRECTORY_TO_COMPRESS=${1%/}
ZIPPED_FILE="$2.zip"
COMPRESS_IGNORE_FILE=("\.git" "*.zip" "*.csv" "*.json" "gulpfile.js" "*.rb" "*.bak" "*.swp" "*.back" "*.merge" "*.txt" "*.sh" "bower_components" "node_modules")
COMPRESS_IGNORE_DIR=("bower_components" "node_modules")
IGNORE_LIST=("*/\.*" "\.* "\/\.*"")
if [[ -n $COMPRESS_IGNORE_FILE ]]; then
for IGNORE_FILES in "${COMPRESS_IGNORE_FILE[#]}"; do
IGNORE_LIST+=("$DIRECTORY_TO_COMPRESS/$IGNORE_FILES/*")
done
for IGNORE_DIR in "${COMPRESS_IGNORE_DIR[#]}"; do
IGNORE_LIST+=("$DIRECTORY_TO_COMPRESS/$IGNORE_DIR/")
done
fi
zip -r "$ZIPPED_FILE" "$DIRECTORY_TO_COMPRESS" -x "${IGNORE_LIST[#]}" # >/dev/null
# echo zip -r "$ZIPPED_FILE" "$DIRECTORY_TO_COMPRESS" -x "${IGNORE_LIST[#]}" # >/dev/null
echo $DIRECTORY_TO_COMPRESS "compressed as" $ZIPPED_FILE.
fi
After a few trial and error, I have managed to fix this problem by changing this line:
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/**\*"
to:
ignorelist="$ignorelist $SYNC_WEB_ROOT_BACKUP_DIR/$ignoredir/***"
Not sure why this worked, but it does :)

Shell Scripting: Print directory names and files with specifics

In my script, I am asking the user to input a directory and then list all the files in that specific directory. What I want to do with that is to make the display a little better in which I would be able to display a "/" if the item in the directory is another directory and if it is an executable file (not an executable directory), print with a **".
This is what I have:
echo “Directory: “
read thing
for var123 in $thing*
do
echo $var123
done
In a directory I have a few folders and a few scripts that have the execute permission. when I run the script I want to say
/folder1/subfolder1/
/folder1/subfolder2/
/folder1/file1*
/folder1/file2*
I am new to this and have no clue what I am doing. Any help will be greatly appreciated.
You might want to check and make sure the user inputs something that ends in a / first.
e.g.
[[ $thing =~ '/'$ ]] || thing="$thing/"
Also check if it exists
e.g.
[[ -d $thing ]] || exit 1
Then for checking if it's a directory use the -d test as above. To check if executable file use -x. So putting that all together, try:
#!/bin/bash
echo “Directory: “
read thing
[[ $thing =~ '/'$ ]] || thing="$thing/"
[[ -d $thing ]] || exit 1
for var123 in "$thing"*
do
if [[ -f $var123 && -x $var123 ]]; then
echo "$var123**"
elif [[ -d $var123 ]]; then
echo "$var123/"
else
echo "$var123"
fi
done
ls -F is your friend here - if you want to do it for the current directory:
ls -F
If you want to do it for all files & subfolders of the current directory:
find * -exec ls -Fd {} \;
... and for a given directory:
echo "Directory: "
read DIR
find $DIR/* -exec ls -Fd {} \;
Edit: ls -F will append a / to directories and a * to executables. If you want ** instead, just use sed to replace them:
find $DIR/* -exec ls -Fd {} \; | sed 's/\*$/&&/'
And this approach works in all shells, not just bash.

How to move a single file with (.JPEG, .JPG, .jpeg, .jpg) extensions) and change the extension to .jpg with Linux bash

I have an inotify wait script that will move a file from one location to another whenever it detects that a file has been uploaded to the source directory.
The challenge I am facing is that i need to retain the basename of the file and convert the following extensions: .JPEG, .JPG, .jpeg to .jpg so that the file is renamed with the .jpg extension only.
Currently I have this:
TARGET="/target"
SRC="/source"
( while [ 1 ]
do inotifywait -m -r -e close_write --format %f -q \
$SRC | while read F
do mv "$SRC/$F" $TARGET
done
done ) &
So I need a way to split out and test for those non standard extensions and move the file with the correct extension. All files not having those 4 extensions just get moved as is.
Thanks!
Dave
if [[ "$F" =~ .JPEG\|jpg\|jpeg\|jpg ]];then
echo mv $F ${F%.*}.jpg
fi
Using extglob option with some parameter expansion:
#! /bin/bash
shopt -s extglob
TARGET=/target
SRC=/source
( while : ; do
inotifywait -m -r -r close_write --format %f -q \
$SRC | while read F ; do
basename=${F##*/} # Remove everything before /
ext=${basename##*.} # Remove everything before .
basename=${basename%.$ext} # Remove .$ext at the end
if [[ $ext == #(JPG|JPEG|jpeg) ]] ; then # Match any of the words
ext=jpg
fi
echo mv "$F" "$TARGET/$basename.$ext"
done
done ) &
Try this format. (Updated)
TARGET="/target"
SRC="/source"
(
while :; do
inotifywait -m -r -e close_write --format %f -q "$SRC" | while IFS= read -r F; do
case "$F" in
*.jpg)
echo mv "$SRC/$F" "$TARGET/" ## Move as is.
;;
*.[jJ][pP][eE][gG]|*.[jJ][pP][gG])
echo mv "$SRC/$F" "$TARGET/${F%.*}.jpg" ## Move with new proper extension.
;;
esac
done
done
) &
Remove echo from the mv commands if you find it correct already. Also it's meant for bash but could also be compatible with other shells. If you get an error with the read command try to remove the -r option.

Renaming a set of files to 001, 002,

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.

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