Bash remove quotes for String argument

Question

I need to execute a command in bash that takes a constructed argument. The string argument needs to be passed without quotations. How do I remove the quotes?

This does not work:

Tracks='bark.mov'
TrackDir='~/soundtracks/'
TrackPath=$TrackDir$Track
omxplayer -o local $TrackPath

This does not work:

omxplayer -o local '~/soundtracks/bark.mov'

This does work:

omxplayer -o local ~/soundtracks/bark.mov

@PatrickHaugh Double quotes also prevent tilde expansion.

John Kugelman
– John Kugelman

2018-06-03 00:17:54 +00:00
Commented Jun 3, 2018 at 0:17 — John Kugelman
– John Kugelman, Commented Jun 3, 2018 at 0:17
TrackDir=~/'soundtracks/'

Cyrus
– Cyrus

2018-06-03 00:37:52 +00:00
Commented Jun 3, 2018 at 0:37 — Cyrus
– Cyrus, Commented Jun 3, 2018 at 0:37

John Kugelman · Accepted Answer · 2018-06-03 00:17:07Z

3

Tilde expansion only works outside of quotes. Get rid of the quotes when defining the variables, but do include quotes when expanding them.

Tracks=bark.mov
TrackDir=~/soundtracks/
TrackPath=$TrackDir$Track
omxplayer -o local "$TrackPath"

answered Jun 3, 2018 at 0:17

John Kugelman

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Comments

chepner · Accepted Answer · 2018-06-03 00:19:20Z

1

You need to leave the ~ unquoted so that the shell can expand it to the user's home directory. Further, it's better practice to add the path separator when you join the two, to make it explicit. Having two adjacent / in the path doesn't hurt.

Tracks=bark.mov
TrackDir=~/soundtracks/
TrackPath=$TrackDir/$Track
omxplayer -o local "$TrackPath"

answered Jun 3, 2018 at 0:19

chepner

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