Modify a path stored in a bash script variable

Question

I have a variable f in a bash script

f=/path/to/a/file.jpg

I'm using the variable as an input argument to a program that requires and input and an output path.

For example the program's usage would look like this

./myprogram -i inputFilePath -o outputFilePath

using my variable, I'm trying to maintain the same basename, change the extension, and put the output file into a sub directory. For example

./myprogram -i /path/to/a/file.jpg -o /path/to/a/new/file.tiff

I'm trying to do that by doing this

./myprogram -i "$f" -o "${f%.jpg}.tiff"

of course this keeps the basename, changes the extension, but doesn't put the file into the new subdirectory.

How can I modify f to to change /path/to/a/file.jpg into /path/to/a/new/file.tiff?

Your goal is unclear. If you want to completely change f, you'll have to reassign it, i.e., by f=/path/to/a/new/file.tiff. if you want to keep some part of f, there are ways just as dirname or basename or your method with parameter expansion. — jeremysprofile
– jeremysprofile, Commented Jun 13, 2018 at 18:37
my goal isn't to modify f persay but to use the path in f to specify the -o path/filename — agf1997
– agf1997, Commented Jun 13, 2018 at 18:38

ErikMD · Accepted Answer · 2018-06-13 18:57:25Z

2

Actually you can do this in several ways:

Using sed as pointed out by @anubhava

Using dirname and basename:

./myprogram -i "$f" -o "$(dirname -- "$f")/new/$(basename -- "$f" .jpg).tiff"

Using only Bash:

./myprogram -i "$f" -o "${f%/*}/new/$(b=${f##*/}; echo -n ${b%.jpg}.tiff)"

Note that unlike the second solution (using dirname/basename) that is more robust, the third solution (in pure Bash) won't work if "$f" does not contain any slash:

$ dirname "file.jpg"
.
$ f="file.jpg"; echo "${f%/*}"
file.jpg

edited Jun 13, 2018 at 18:57

answered Jun 13, 2018 at 18:44

ErikMD

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anubhava · Accepted Answer · 2018-06-13 18:39:59Z

1

You may use this sed:

s='/path/to/a/file.jpg'
sed -E 's~(.*/)([^.]+)\.jpg$~\1new/\2.tiff~' <<< "$s"

/path/to/a/new/file.tiff

answered Jun 13, 2018 at 18:39

anubhava

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chepner · Accepted Answer · 2018-06-13 19:06:48Z

1

If you're on a system that supports the basename and dirnamecommands you could use a simple wrapper function eg:

$ type newSubDir
newSubDir is a function
newSubDir ()
{
    oldPath=$(dirname "${1}");
    fileName=$(basename "${1}");
    newPath="${oldPath}/${2}/${fileName}";
    echo "${newPath}"
}

$ newSubDir /path/to/a/file.jpg new
/path/to/a/new/file.jpg

If your system doesn't have those, you can accomplish the same thing using string manipulation:

$ file="/path/to/a/file.jpg"

$ echo "${file%/*}"
/path/to/a

$ echo "${file##*/}"
file.jpg

edited Jun 13, 2018 at 19:06

chepner

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answered Jun 13, 2018 at 18:46

drldcsta

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