This question already has answers here:
Get just the filename from a path in a Bash script [duplicate]
(6 answers)
Closed 4 years ago.
Let's say I have a string which represents the full path of a file:
full_path='./aa/bb/cc/tt.txt'.
How can I extract only the file name tt.txt?
Please don't tell me to use echo $full_path | cut -d'/' -f 5.
Because the file may be located in a deeper or shallower folder.
The number, 5, cannot be applied in all cases.
if you are comfortable with python then, you can use the following piece of code.
full_path = "path to your file name"
filename = full_path.split('/')[-1]
print(filename)
Use the parameter expansion functionality.
full_path='./aa/bb/cc/tt.txt'
echo ${full_path%/*}
That will give you the output
./aa/bb/cc
And will work any number of directory levels deep, as it will give you everything up to the final "/".
Edit: Parameter expansion is very useful, so here's a quick link for you that gives a good overview of what is possible: http://wiki.bash-hackers.org/syntax/pe
You can use this construct
echo "$full_path" | sed 's/\/.*\///'
Related
This question already has answers here:
Usage of :- (colon dash) in bash
(2 answers)
Closed 1 year ago.
I ran across the following Bash function which was suggested as a useful alias to add to the .bashrc file. It lists the last 13 files in a directory that were modified.
I don't understand what is being done with the ${1:-.} argument, though. It looks like some kind of substring extraction, but I couldn't find the meaning of -. in the Advanced Bash Scripting Manual.
I tried the command in a few directories and didn't notice any difference between the output when I removed this argument. My guess is that it's there to prevent an error when encountering some specific type of file or file name. What is it doing? And what is the purpose of including it in the command?
function lst()
{
ls -lashtg ${1:-.} | head -13
}
$1 is the first command line argument. Im sure you know.
${1:-.} simply puts a . when no first line argument is given.
Thus
lst
Translates to
ls -lashtg . | head -13
It would workout without the substitution I guess. But I also guess that this is just there as a best practice
This question already has answers here:
Rename multiple files based on pattern in Unix
(24 answers)
Closed 2 years ago.
I have a lot of file with the same format (???_ideal.sdf) name and I need to rename all (???.sdf) removing form the name "ideal". For example:
Files:
002_ideal.sdf
ERT_ideal.sdf
234_ideal.sdf
sCX_idel.sdf
New Files:
002.sdf
ERT.sdf
234.sdf
SCX.sdf
I thought to using a loop but I don’t know how to indicate that in the new file name should be removed "individual".
For example:
for file in ???_individual.sdf; mv $file what?
With your shown samples, could you please try following. This will only print the rename command on terminal(for safer side, check and make sure if commands are fine and looking ok to you first before actually renaming files), to perform actual rename remove echo from following.
for file in *_ideal.sdf
do
echo mv "$file" "${file/_ideal/}"
done
Based on this answer you can also write it as one line with:
rename 's/^(.*)_ideal.png/$1.png/s' **/**
First parameter replaces the the _ideal.png, second one means all files.
This question already has answers here:
Command not found error in Bash variable assignment
(5 answers)
Closed 2 years ago.
i am trying to give an file as input to me my shell script:
#!/bin/bash
file ="$1"
externalprogram "$file"
echo 'unixcommand file '
i am trying to give the path to my file but it says always
cannot open `=/home/username/documents/file' (No such file or directory)
my path is this /home/username/Documents/file
i do this in terminal : ./myscript.sh /home/username/Documents/file
can someone help me please?
When you say
file ="$1"
with a space after "file", you're running something called file with =$1 as an argument. There probably actually is a utility called file. If you want to assign $1 to a variable called file, you don't need the space:
file="$1"
there shouldn't be a space before = in the second line.
file=$1 should be good enough.
Check what shellcheck says about your code:
^-- SC1068: Don't put spaces around the = in assignments (or
quote to make it literal).
You can read more about SC1068 case on its Github
page.
#!/bin/bash
file=$1
code $file
echo "aberto o arquivo ${file} no vscode"
I made this code snippet to demonstrate, I pass a path and it opens the file in vscode
This question already has an answer here:
In bash, how do I expand a wildcard while it's inside double quotes?
(1 answer)
Closed 4 years ago.
I am trying to add a line to a specific file which matches a pattern in its name.
e.g. I am trying to append text STATUS PASSED to a file whose name contains 2018_09_26_04_51_30.
date="2018_09_26_04_51_30"
echo "STATUS PASSED" >> "/test_dir/*$date*.txt"
Above said commands are creating new file named *2018_09_26_04_51_30*.txt which is not serving my purpose!
Let's assume that there are several other files with *.txt extension, but none of these files contains $date in their names.
The test_dir directory contains:
test1-2018_09_26_04_50_48.txt
test2-2018_09_26_04_50_56.txt
test3-2018_09_26_04_51_03.txt
test3-2018_09_26_04_51_30_51S.txt
So, here file test3-2018_09_26_04_51_30_51S.txt is unique.
P.S. I have to execute this script in both Linux and AIX.
Any help will be appreciated!
Try like this,
date="2018_09_26_04_51_30"
#cd /test_dir
echo "STATUS PASSED" >> $(ls /test_dir/*$date*.txt)
This question already has answers here:
Getting the parent of a directory in Bash
(13 answers)
Closed 1 year ago.
In my script I need the directory of the file I am working with. For example, the file="stuff/backup/file.zip". I need a way to get the string "stuff/backup/" from the variable $file.
dirname $file
is what you are looking for
dirname $file
will output
stuff/backup
which is the opposite of basename:
basename $file
would output
file.zip
Using ${file%/*} like suggested by Urvin/LuFFy is technically better since you won't rely on an external command. To get the basename in the same way you could do ${file##*/}. It's unnecessary to use an external command unless you need to.
file="/stuff/backup/file.zip"
filename=${1##*/} # file.zip
directory=${1%/*} # /stuff/backup
It would also be fully POSIX compliant this way. Hope it helps! :-)
For getting directorypath from the filepath:
file="stuff/backup/file.zip"
dirPath=${file%/*}/
echo ${dirPath}
Simply use $ dirname /home/~username/stuff/backup/file.zip
It will return /home/~username/stuff/backup/