How to change which program is executed by default

executablepath

Surely this is an easy question, that I just don't know how to search for, but if I have two identically named executable files, one in /usr and one in /usr/local (for example), how can I change which one is executed by default without specifying the path, as in /usr/local/file, if $ which returns /usr?

Centos6.4 btw.

Best Answer

It's not always as simple as "which comes first in $PATH;" see https://superuser.com/questions/358695/how-does-linux-decide-which-executable-im-trying-to-run. For a quick fix, set an alias in .bashrc (assuming you're using bash...)

alias gorgonzola='usr/local/gorgonzola'

Note: white space is not allowed around the "=" sign.

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bash – How to Clear Bash’s Cache of Paths to Executables

bash does cache the full path to a command. You can verify that the command you are trying to execute is hashed with the type command:

$ type svnsync
svnsync is hashed (/usr/local/bin/svnsync)

To clear the entire cache:

$ hash -r

Or just one entry:

$ hash -d svnsync

For additional information, consult help hash and man bash.

Bash – Default preference of executable over built-ins with the same name

My question is, is there a way to make an executable file in $PATH have preference over an shell builtin, by executing only, e.g. cd without builtin or command?

You can use the builtin enable to disable/enable a builtin. Say:

enable -n cd

to disable the builtin cd. Say enable cd to enable the builtin.

The following would give an example of switching between the builtin and command:

$ type cd
cd is a shell builtin
$ enable -n cd
$ type cd
-bash: type: cd: not found
$ enable cd
$ type kill
kill is a shell builtin
$ enable -n kill
$ type kill
kill is /bin/kill
$ enable kill
$ type kill
kill is a shell builtin

Best Answer

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bash – How to Clear Bash’s Cache of Paths to Executables

Bash – Default preference of executable over built-ins with the same name

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