Bash – How to load .bashrc from “bash -c”

bash

Trying to run a function defined in my .bashrc using "bash -c ". I end up with the error "command not found". How do I get "bash -c" to load my init file?

Best Answer

You can make it into an interactive shell with -i, then your ~/.bashrc will be read:

bash -i -c "echo \$EDITOR"

The other thing you can do is source the file explicit. If you have /var/tmp/test with content:

export XXX=123

and you do

bash -c "source /var/tmp/test; echo \$XXX"

you will get 123 echoed.

Related Solutions

Bash – Using $? in an if statement

Add a space after the [, and another before ]:

function foo {
   (cd "$FOOBAR";
   <some command>
   if [ "$?" -ne 0 ]
   then
      echo "Nope!"
   else
      echo "OK!"
   fi
   )
}

[ is a shell builtin, it is a command just like echo, read, expr... it needs a space after it, and requires a matching ].

Writing [ "$?" -ne 0 ] is actually invoking [ and giving it 4 parameters: $?, -ne, 0, and ].

Note: the fact that you are getting an error saying [1: command not found means that $? was having the value of 1.

Best Answer

Related Solutions

Bash – Using $? in an if statement

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