I am trying to automate a process which take a file type NIFTI, preprocess it, and places the new processed file in an output folder.
deepbrain-extractor -i <input-dir> -o <output-dir>
I wrote this bash script to automate this process for all files in a directory:
for file in path/*.nii
do
deepbrain-extractor -i $file -o path/newfiles
done
The problem is every time the code runs, it overwrites the old files (since all automatically get the same name). Is there a way to prevent that?
Thanks for the suggestions and comments!
seems like using this code gives me what I want.
deepbrain-extractor -i "$file" -o "${file%.nii}.out"
Thanks a lot!!
A common way to get a unique file name is to use the current date and time.
You can combine this with the input file name to ensure that output files are never overwritten.
if [ ! -f "$FILE" ]; then
echo "$FILE does not exist."
for file in path/*.nii
do
deepbrain-extractor -i $file -o path/newfiles_${file##*/}_$(date +"%Y-%m-%d_%H-%M-%S")
done
fi
so the output directory has a name like newfiles_input_2020-11-13_12-34-56
Also see How to create one output file for each file passed to a loop in bash?
Related
I'm looking to check if a variable file is in another directory, and if it is, stop the script from running any farther. So far I have this:
#! /bin/bash
for file in /directory/of/variable/file/*.cp;
do
test -f /directory/to/be/checked/$file;
echo $?
done
I ran an echo of $file and see that it includes the full path, which would explain why my test doesn't see the file, but I am at a loss for how to move forward so that I can check.
Any help would be greatly appreciated!
Thanks
I think you want
#! /bin/bash
for file in /directory/of/variable/file/*.cp ; do
newFile="${file##*/}"
if test -f /directory/to/be/checked/"$newFile" ; then
echo "/directory/to/be/checked/$newFile already exists, updating ..."
else
echo "/directory/to/be/checked/$newFile not found, copying ..."
fi
cp -i "$file" /directory/to/be/checked/"$newFile"
done
Note that you can replace cp -i with mv -i and move the file, leaving no file left behind in /directory/of/variable/file/.
The -i option means interrogate (I think), meaning if the file is already there, it will ask you overwrite /directory/to/be/checked/"$newFile" (or similar) to which you must reply y. This will only happen if the file already exists in the new location.
IHTH
The command basename will give you just the file (or directory) without the rest of the path.
#! /bin/bash
for file in /directory/of/variable/file/*.cp;
do
test -f /directory/to/be/checked/$(basename $file);
echo $?
done
I have a bunch of .sh files in a directory. I need to find out how many stored procedures each one of them calls. What would be the best way to do that?
I'm REALLY new with bash scripts so this is all very new to me. From what I looked online I hacked up a starting point (I think) but I have no idea how I would open each file, and find "something.sql" in it and then out put number of times that was found in each file.
Here's what I have:
#!/bin/sh
for i in 'ls *.sh'
echo -e "\n **** START****"
do
echo -e " \n Filename: $i"
done
echo -e "\n **** END ****"
done
Thanks for any help!
Try this:
grep -nc sql *.sh
See how that moves you. You can add -i if you name sql files in as file.SQL too. Or if they all have a .sql extension.
grep -nc '\.sql' *.sh
For you comment you added, try this:
for i in *.sh
grep -Hc '\.sql' $i
grep '\.sql' $i
done
I have been using an image optimizer for my websites and when I do this, it gives me files with -compressor at the end of it.
input: filename.jpg
output: filename-compressor.jpg
I need help in creating a batch file or a command script that I can just place these files into a folder and it will loop through all of these and change the names of these for me so that I don't have to go through them one by one.
mkdir -p compressors
mv *-compressor.jpg compressors/
cd compressors
for i in *-compressor.jpg; do j=${i%%\-compressor.jpg}.jpg; mv "$i" "$j"; done
src_dir="/export/home/destination"
list_file="client_list_file.txt"
file=".csv"
echo "src directory="$src_dir
echo "list_file="$list_file
echo "file="$file
cd /export/home/destination
touch $list_file
x=`ls *$file | sort >$list_file`
if [ -s $list_file ]
then
echo "List File is available, archiving now"
y=`tar -cvf mystuff.tar $list_file`
else
echo "List File is not available"
fi
The above script is working fine and it's supposed to create a list file of all .csv files and tar's it.
However I am trying to do it from a different directory while running the script, so it should go to the destination directory and makes a list file with all the .csv in destination directory and make a .tar from the list file(i.e archive the list file)
So i am not sure what to change
there are a lot of tricks in filename handling. the one thing you should know is file naming under POSIX sucks. commands like ls or find may not return the expected result(but 99% of the time they will). so here is what you have to do to get the list of files truely:
for file in $src_dir/*.csv; do
echo `basename $file` >> $src_dir/$list_file
done
tar cvf $src_dir/mystuff.tar $src_dir/$list_file
maybe you should learn bash in a serious manner and try to google first before you asking question in SO next time.
http://www.gnu.org/software/bash/manual/html_node/index.html#SEC_Contents
http://tldp.org/HOWTO/Bash-Prog-Intro-HOWTO.html
I'm new in shell script and I am trying to figure out a way to write a script that copies all the files in the current directory to a directory specified from a .txt file and if there are matching names, it adds the current date in the form of FileName_YYYYMMDDmmss to the name of the file being copied to prevent overwritting.
Can someone help me out?
I saw thinking something along the lines of
#!/bin/bash
source=$pwd #I dont know wheter this actually makes sense I just want to
#say that my source directory is the one that I am in right now
destination=$1 #As I said I want to read the destination off of the .txt file
for i in $source #I just pseudo coded this part because I didn't figure it out.
do
if(file name exists)
then
copy by changing name
else
copy
fi
done
the problem is I have no idea how to check whether the name exist and copy and rename at the same time.
Thanks
How about this? I am supposing that the target directory is in the
file new_dir.txt.
#!/bin/bash
new_dir=$(cat new_dir.txt)
now=$(date +"%Y%m%d%M%S")
if [ ! -d $new_dir ]; then
echo "$new_dir doesn't exist" >&2
exit 1
fi
ls | while read ls_entry
do
if [ ! -f $ls_entry ]; then
continue
fi
if [ -f $new_dir/$ls_entry ]; then
cp $ls_entry $new_dir/$ls_entry\_$now
else
cp $ls_entry $new_dir/$ls_entry
fi
done
I guess this what you are looking for :
#!/bin/bash
dir=$(cat a.txt)
for i in $(ls -l|grep -v "^[dt]"|awk '{print $9}')
do
cp $i $dir/$i"_"$(date +%Y%m%d%H%M%S)
done
I assumed that a.txt contains only the name of the destination directory. If there are other entries, you should add some filter to the first statement(using grep or awk).
NB: I used full time stamp(YYYYMMDDHHmmss) instead of your YYYYMMDDmmss as it doesn't seem logical.