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I have a Bash script with the following lines of code:

echo "current directory is" $PWD

echo $PWD

# put current directory into a variable for use later in script
originalDirectory=$PWD

echo "contents of original directory variable:" originalDirectory

And the resultant output is this: enter image description here

The first 2 lines of output make sense but the 3rd one does not. I expected it to output
contents of original directory variable: /media/sf_code/scripts
but it is just printing the name of the variable. Why doesn't it display the contents of the originalDirectory variable after the colon in that string?

5

The last line should be

echo "contents of original directory variable:" $originalDirectory

Without the $ it is text, not a variable content. So you get the text.

So it is an expected output ;-)

  • 1
    For echo the variable name wouldn't need double quoutes, but for other context it should, in case there are spaces in the directory. – guillermo chamorro Dec 14 '19 at 16:04

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