bash - How to use DATE command with an external variable? -

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i'm working on time convertion script. supposed this:

echo $(( ($(date -d '00:10:2.00' +%s) - $(date -d 0 +%s) ) ))

this line working fine giving me result

602

but want put the first part of date string (00:10:2.00) under a) command line argument read $1 like:

echo $(( ($(date -d '$1' +%s) - $(date -d 0 +%s) ) ))

or variable:

echo $(( ($(date -d '$myvariable' +%s) - $(date -d 0 +%s) ) ))

when i'm trying this:

foo="00:10:2.00" echo $foo echo $(( ($(date -d '$foo' +%s) - $(date -d 0 +%s) ) ))

all is:

00:10:2.00 date: invalid date `$foo' -1417042800

so it's echoing aint working time command...

variables expanded inside double quotes, they're not expanded inside single quotes. use

date -d "$1" +%s

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