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What I am trying to do in bash is write a script, foo, that when I run $(foo), it changes the directory and runs a script using source. The contents of foo:

echo cd bar
echo "&&"
echo source baz

When I run $(foo), it changes directory to "bar", but it does not run source baz. Is there any way to accomplish this?

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2  
Plus, why would you want to do that? This seems a little contrived and I'm sure there are better ways to achieve what you need. –  slhck Feb 6 '13 at 21:42
    
@slhck I can't affect the shell from inside a script or program. The only ways to affect the shell are with command substitution or source, so I'm trying to find a way to do that. –  Kevin Marsolais Feb 6 '13 at 22:13
    
Why are you storing the commands to execute in a variable in the first place? –  chepner Feb 8 '13 at 13:53

2 Answers 2

up vote 1 down vote accepted

You need to eval the output:

eval $(foo)

Otherwise, &&(or any other similar language element) is treated as if it were wrapped in quotation marks: a regular argument to cd.

$ echo $(foo)
cd bar && source baz

Skipping the leading echo, this would execute cd with the four arguments bar && source baz, silently dropping all but the first.

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Note that, depending on your application, this might not be a really good idea. –  Daniel Beck Feb 6 '13 at 22:27
    
Seems to work, if not the most elegant solution. Thanks. –  Kevin Marsolais Feb 6 '13 at 22:48

You can use aliases (see also: man bash) to do what your trying to do.

alias foo='cd bar && source baz'

will change directory to bar and source baz when you execute foo.

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