command command in bash: Run command with arguments ignoring any shell function named command.
The '-p' option means to use a default value for $PATH that is guaranteed to find all of the standard utilities.
What exactly is the default PATH mentioned here? When I define
export PATH="/home/ozgur/":$PATH, don't I add a new PATH path over the default value?
## For Example ~$ export PATH="/home/ozgur/":$PATH ~$ echo $PATH /home/ozgur/:/usr/local/bin:/usr/bin:/bin:/usr/local/games:/usr/games ~$ script.sh hey, i am working ! ~$ command -p script.sh hey, i am working !
When I used the "-p" option of the
command command, I would expect it to ignore the new PATH path I had defined, but that didn't happen. What exactly am I missing here? What is the point of using the "-p" option if changes to the user's PATH path are not overridden with the "-p" option?