I have set few aliases in .bashrc file. I need those aliases most of the time, but sometimes I need to run those command without options set in that particular alias.

How to not execute alias command?

0

3 Answers

Run the command with a leading \, an answer in examples: ;)

% ls
bar foo
% alias ls="ls -laog"
% ls
total 4292
drwxrwxr-x 4 4329472 Nov 5 15:06 .
drwx------ 95 28672 Nov 5 15:15 ..
-rw-rw-r-- 1 0 Nov 5 15:06 bar
drwxrwxr-x 2 4096 Nov 5 15:06 foo
drwxrwxr-x 2 4096 Okt 2 14:29 .foo
-rw-rw-r-- 1 191 Feb 25 2015 .htaccess
% \ls
bar foo

---

Slightly longer but also possible:

command ls

3

You can use shell builtin command to escape aliases (and functions):

command alias_name

For example:

command ls

will run /bin/ls binary , not any alias defined as ls.

An alternative is to use quotes:

"alias_name"

or

'alias_name'

For example:

"ls"

or

'ls'

these again will run the /bin/ls binary, ignoring any alias ls.

5

Alternatively, you could specify full path to command. For instance, my ls command is aliased to ls='ls --color -F'

What one can do is either call /bin/ls or $(which ls) (for those of us who are lazy to type full path).

Example : calling original command with flags $(which ls) -l

2