exporting variables which are variable values in shell

Question

When i try to do this...

$ export my_path='/my/path\ with\ spaces/file'
$ echo $my_path`
/my/path\ with\ spaces/file

It works.

But i want to do this..

$ echo $my_variable
my_path='/my/path\ with\ spaces/file'
$ export $my_variable
-bash: export: `with\': not a valid identifier
-bash: export: `spaces/file'': not a valid identifier

This is the error iam getting. Is there a way to treat the value of the variable for exporting.. [[ NOTE: if the path did not have a space, it works perfectly!! ]]

hek2mgl · Accepted Answer · 2013-05-06 09:38:40Z

3

Don't use the $. Otherwise the shell would expand the variable's value before passing it to export. The following will work:

export my_variable

In comments, it points out that you wan't to store the variable assignment itself as a variable and then pass it to export. Here comes an example:

var='foo=bar'
export "$var"

# check if it succeeded 
export | grep foo # output: declare -x foo="bar"

edited May 6, 2013 at 9:38

answered May 6, 2013 at 9:02

hek2mgl

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5 Comments

user1019830 Over a year ago

This works because export will expand the value of its arguments. When you say export $foo then foo will be expanded before being passed to export and if foo contains spaces, export will think it's given several arguments that are names of shell variables.

raj Over a year ago

actually, my end goal is to export the "my_path" variable. Not the "my_variable" itself.

hek2mgl Over a year ago

Something like this? : $var='$foo=bar'; export "$var"

raj Over a year ago

Not working man.. :( bash: export: '$foo=bar': not a valid identifier

hek2mgl Over a year ago

var='foo=bar'; export "$var"

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