I need to create a text file in each subdirectory of all files in the list.
For example, subdirectory1 would contain a list of all of its files as a .txt and subdirectory2 would also contain a list of all of subdirectory2 files as a .txt.
I have tried
#!/bin/bash
for X in "$directory" *
do
if [ -d "$X" ];
then
cd "$X"
files="$(ls)"
echo "$files" >> filesNames.txt
fi
done
However this did not generate anything. I absolutely need it as a shell script because it will be part of a pipeline script, but I cannot seem to get it to work.
Here is the adjusted script giving me the no such file or directory comment. I know that the folder exists and have used it in commands that are run before this command.
#!/bin/bash
#Retrieve the base directory path
baseDir=$(dirname "$ini")
#Retrieve the reference genome path
ref=$(dirname "$genome")
#Create required directory structure
tested="$baseDir/tested"
MarkDups1="$baseDir/MarkDups1"
#don't create if already exists
[[ -d "tested" ]] || mkdir "$tested"
[[ -d "MarkDups1" ]] || mkdir "$MarkDups1"
#create a text file with all sorted and indexed bam files paths
#!/bin/bash
for x in $MarkDups1/*/;
do
(cd "$x"; ls > filesNames.txt)
done
The sequence to iterate over should be "$directory"/*/.
for x in "$directory"/*/; do
(cd "$x"
files=(*)
printf '%s\n' "${files[#]}" > filesNames.txt
)
done
Related
sorry for that odd title. I didn't know how to word it the right way.
I'm trying to write a script to filter my wiki files to those got directories with the same name and the ones without. I'll elaborate further.
here is my file system:
what I need to do is print a list of those files which have directories in their name and another one of those without.
So my ultimate goal is getting:
with dirs:
Docs
Eng
Python
RHEL
To_do_list
articals
without dirs:
orphan.txt
orphan2.txt
orphan3.txt
I managed to get those files with dirs. Here is me code:
getname () {
file=$( basename "$1" )
file2=${file%%.*}
echo $file2
}
for d in Mywiki/* ; do
if [[ -f $d ]]; then
file=$(getname $d)
for x in Mywiki/* ; do
dir=$(getname $x)
if [[ -d $x ]] && [ $dir == $file ]; then
echo $dir
fi
done
fi
done
but stuck with getting those without. if this is the wrong way of doing this please clarify the right one.
any help appreciated. Thanks.
Here's a quick attempt.
for file in Mywiki/*.txt; do
nodir=${file##*/}
test -d "${file%.txt}" && printf "%s\n" "$nodir" >&3 || printf "%s\n" "$nodir"
done >with 3>without
This shamelessly uses standard output for the non-orphans. Maybe more robustly open another separate file descriptor for that.
Also notice how everything needs to be quoted unless you specifically require the shell to do whitespace tokenization and wildcard expansion on the value of a token. Here's the scoop on that.
That may not be the most efficient way of doing it, but you could take all files, remove the extension, and the check if there isn't a directory with that name.
Like this (untested code):
for file in Mywiki/* ; do
if [ -f "$d" ]; then
dirname=$(getname "$d")
if [ ! -d "Mywiki/$dirname" ]; then
echo "$file"
fi
fi
done
To List all the files in current dir
list1=`ls -p | grep -v /`
To List all the files in current dir without extension
list2=`ls -p | grep -v / | sed 's/\.[a-z]*//g'`
To List all the directories in current dir
list3=`ls -d */ | sed -e "s/\///g"`
Now you can get the desired directory listing using intersection of list2 and list3. Intersection of two lists 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.
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 :)
I have a series of directories that are only different by a numerical tag.
arr=(0 1 2 3)
i=0
while [ $i -le ${arr}]
do
dir="~Documents/seed"
dir+=${arr[i]}
echo $dir #works
cd dir #directory not found
#do other things#
done
Is it possible to do this?
This might be easier:
#!/bin/bash
for d in ~/Dcouments/seed*
do
if [ -d "$d" ]; then
echo $d
fi
done
Note:
You have tarfiles in ~/Documents too (with names that also match the wildcard), so I have added an if statement that checks if it is a directory or a file and only reacts to directories.
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.