Question

The script below should open an entry box whose initial entry-text is the value of $filename, but the zenity section in the third line ignores that variable and I end up with no entry-text.

By the way, if I just replace filename by whoami, my username will appear as expected. I’m not able to get a result using my own variables, though. That’s the problem.

#!/bin/bash
filename="duck"
sh -c 'chromium-browser "https://duckduckgo.com/?q=%5C$(zenity --entry --text "I am feeling ducky" --entry-text $(filename))"'

And if I removed the sh -c stuff, to end up with the code below, my variable would work with zenity.

#!/bin/bash
filename="duck"
zenity --entry --text "I am feeling ducky" --entry-text $filename

Any clues as to why? I tried playing with different combination of quotes but my efforts were futile.

Answer 1

The notation $(filename) runs a command ‘filename’.

To expand a variable, use ${filename}.

Note, too, that unless $filename is an exported variable, the sub-shell (the one that runs the -c option) will not see the value of the variable (because of the single quotes).

To fix that, you’d have to use double quotes – rather carefully:

#!/bin/bash
filename="duck"
sh -c "chromium-browser \"https://duckduckgo.com/?q=%5C$(zenity --entry --text \"I am feeling ducky\" --entry-text ${filename})\""