This question already has answers here:
How do I know the script file name in a Bash script?
(25 answers)
Closed 2 years ago.
I want to save a copy of the bash file each time I run it. It should be saved to the output dictionary.
I am doing it like this in the mytrainrtest.sh file:
mkdir -p "${EXP_DIR}/train"
cp "${WORK_DIR}"/mytrainrtest.sh "${EXP_DIR}"/.
Now I have much more bash files with name my****** as copies of the upper one, each with different names.
How can I write the line, so the bash file will recognize its name to copy itself?
Use the $0 special variable which contains the name of the currently executing script.
cp "$0" "$exp_dir"/
Script name(path) stored in a special var $0
#!/bin/bash
echo $0
$ ./test
./test
Related
This question already has answers here:
Dynamic variable names in Bash
(19 answers)
Closed 7 months ago.
I have a bash script (list.sh) that lists files in a folder and assigns each filename to a bash variable:
#/bin/bash
c=0
for file in *; do
varr$c="$file";
c=$((c+1));
done
When I call this from a bash terminal with:
source list.sh
I get:
bash: varr0=07 Get the Balance Right!.mp3: command not found...
bash: varr1=190731_10150450783260347_1948451_n.jpg: command not found...
bash: varr2=199828_10150450907505347_7125763_n.jpg: command not found...
bash: varr3=2022-07-31_19-30.png: command not found...
bash: varr4=2022-08-02_12-06.png: command not found...
bash: varr5=246915_10152020928305567_1284271814_n.jpg: command not found...
I don't know how to put quotes around the file text itself, so that each varr(x) creates itself as a variable in the parent bash script, ie:
varr0="07 Get the Balance Right!.mp3"
It's not a "quote around the text" issue, it's the variable declaration varr$c that's not working. You should be using declare instead.
This script solves your problem :
#/bin/bash
c=0
for file in *; do
declare varr$c=$file;
c=$((c+1));
done
You can use the keyword declare as in
$ n=1
$ declare var_$n=20
$ echo $var_1
20
https://www.linuxshelltips.com/declare-command-in-linux/
Try this script:
c=0
for file in *; do
printf -v var$((c++)) '%s' "$file"
done
# list variables starting with var and their values
for v in ${!var*}; do
printf '%s=%s\n' "$v" "${!v}"
done
Though using an array must have been preferred over this method.
This question already has answers here:
Do not show results if directory is empty using Bash
(3 answers)
Closed 4 years ago.
So I have to find all the files in the directory that start with the letter a, and list them out. This is pretty easy by doing
cd some_directory
for file in a*; do
echo "$file"
done
However I want that if there are no files present that match a*, then the for loop will not run at all. Currently, if this is the case then the shell will echo
a*
Is there a way to do this? Thank you
Your text is opposite of your title, in my answer below I've assumed the text is your intention and your title is incorrect:
globs can be made to act like this with the bash shell option "nullglob":
shopt -s nullglob
An alternative is to use find and ignore errors by piping stderr to /dev/null
for file in $(find a* 2>/dev/null); do
echo "$file"
done
This question already has answers here:
How to cat multiple files from a list of files in Bash?
(6 answers)
Closed 4 years ago.
How should i concatenate multiple text files get by arguments in terminal using a script in Bash?
#!/bin/bash
while read $1
do
cat $1 > cat.txt
done
I tried that example but it is not working.
You should use '>>' to concatenate ('>' will create a new file each time):
for file in "$#"
do
cat $file >> result
done
This question already has answers here:
Bash - Initialize variable to output of command "permission denied"
(2 answers)
Closed 3 years ago.
I am trying to run this simple script in which I loop through a series of files in a directory and want to create a variable with 'cut', to extract part of the name of the files.
I get a permission denied error and cannot figure out why.
Below is my script.
FILES=./data/*
for f in $FILES
do
NEWNAME=$($f|cut -c3-12)
echo $NEWNAME
done
The ultimate goal is to create one directory per file for downstream processing, with mkdir /path/to/new/directory/$NEWNAME.
$(stuff) executes stuff as a command and substitutes its output back into the command line. So when you do:
$($f|cut -c3-12)
it tries to execute $f as a command, and pipes its output to cut. You get an error because the file in $f doesn't have execute permission.
If you're trying to cut the contents of the variable $f, you need to echo it:
NEWNAME=$(echo "$f" | cut -c3-12)
But there's no need to use cut for this, bash has a built-in parameter expansion operator to select a substring:
NEWNAME=${f:2:10}
This question already has answers here:
When to wrap quotes around a shell variable?
(5 answers)
Closed 1 year ago.
I'm new to bash and I am trying to create a script that should find an archive in a given directory. $1 is the name of archive.
When the given path is ./1/ar.tgz the script works. But when path is ../data 1/01_text.tgz I have the following problem:
dirname: extra operand "1/01_text.tgz"
and then No such file or directory.
Here is my code fragment:
VAR=$1
DIR=$(dirname ${VAR})
cd $DIR
What am I doing wrong?
Ahmed's answer is right, but you also need to enclose VAR in double quotes. The correct code fragment is:
VAR=$1
DIR=$(dirname "$VAR")
cd "$DIR"
The space is causing the problem: cd $DIR gets expanded to cd ../data 1/01_text.tgz and cd doesn't know what to make of the third "argument". Add quotes around the directory: cd "$DIR".