Resolved: Replacing the last line in Bash

Question:

I have a shell script that’s auto generated by a software and it looks something like:

#!/bin/sh

# The directory of this script is the expanded absolute path of the "$qt_prefix/bin" directory.
script_dir_path=`dirname $0`
script_dir_path=`(cd "$script_dir_path"; /bin/pwd)`

/home/akshay/Qt/6.2.4/gcc_64/bin/qmake -qtconf "$script_dir_path/target_qt.conf" $*

I want to replace the last line’s first three separators to:

#!/bin/sh

# The directory of this script is the expanded absolute path of the "$qt_prefix/bin" directory.
script_dir_path=`dirname $0`
script_dir_path=`(cd "$script_dir_path"; /bin/pwd)`

/opt/qt/6.2.4/gcc_64/binqmake -qtconf "$script_dir_path/target_qt.conf" $*

I tried getting the last line by doing:

qmake_path=$(tail -n 1 $(find ${QT_ROOT}/android* -maxdepth 0 -type d | sort -r | head -1)/bin/qmake)

Then I used read to split it by space, I don’t know how to get the first index, in this case the path. Any help on this is appreciated.

Answer:

If it is just the short script you show, you can handle the change with sed, e.g.,

sed -i 's/^\/home\/akshay\/Qt/\/opt\/qt/' script.sh

Where -i will edit in-place and the normal substitution of s/find/replace/ is used to locate a line beginning with ^\/home\/akshay\/Qt (e.g. "/home/akshaw/Qt") and replace as \/opt\/qt (e.g. "/opt/qt")
Example Use/Output
Showing the result of the substitution output to stdout you would have:

$ sed 's/^\/home\/akshay\/Qt/\/opt\/qt/' script.sh
#!/bin/sh

# The directory of this script is the expanded absolute path of the "$qt_prefix/bin" directory.
script_dir_path=`dirname $0`
script_dir_path=`(cd "$script_dir_path"; /bin/pwd)`

/opt/qt/6.2.4/gcc_64/bin/qmake -qtconf "$script_dir_path/target_qt.conf" $*

note: you can also use alternate delimiters to avoid having to escape '/', e.g.

sed 's#^/home/akshay/Qt#/opt/qt#' script.sh

That may be easier on the eyes.
Replacing 3-Words Generically
If the text of the first three words in "/one/two/three/..." can change and need to be replaced generically, you can simply use a character list to require one-or-more characters not '/' followed by '/' using [^/]+/, you can do:

sed -E -i 's#^/[^/]+/[^/]+/[^/]+/#/opt/qt/#' script.sh

The -E option is also used to specify “extended REGEX” syntax making the '+' (one-or-more) repetition character available.
With the change, it doesn’t matter what the first 3-words are (so long as there is a '/' after the 3rd word, they will be replaced with "/opt/qt/".

If you have better answer, please add a comment about this, thank you!