Rename using shell variable substitution

[jpoet]

Hi,

I am trying to create a bash script to rename my files.  It will take a couple of arguments which will be used as part of the file name.  I have tried this:

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#!/bin/bash

TRIP=$1

PLACE=$2

if [ "X$TRIP" = "X" ]
then
    echo "TRIP must be provided"
    exit 1
fi

if [ "X$PLACE" = "X" ]
then
    echo "PLACE must be provided"
    exit 1
fi

FORMAT="/home/john/Pictures/${TRIP}-%Y/%Y%m%d-%H%M%S"
exiftool -r -o . -d $FORMAT '-filename<${DateTimeOriginal}${SubSecTimeOriginal}-${fileindex;$_=sprintf("%.5d",$_)}%+2c-${PLACE}-${artist;tr/ /_/}.%e'  .


Using ${TRIP} as part of the format works.  Having ${PLACE} as part of the 'filename<..." fails with:

Warning: [minor] Tag 'PLACE' not defined - <filename>
Warning: No writable tags set from <filename>
    0 image files updated

I tried escaping ${PLACE} in every way I could think of, but cannot get it to work.  Can someone please tell me the correct format?

Thanks!

[Phil Harvey]

Shell scripts interpolate variables in double quotes, not single quotes.  So you need the to use double quotes for the shell variable.  Maybe something like this:

exiftool -r -o . -d $FORMAT '-filename<${DateTimeOriginal}${SubSecTimeOriginal}-${fileindex;$_=sprintf("%.5d",$_)}%+2c-'"${PLACE}"'-${artist;tr/ /_/}.%e'  .

- Phil

[jpoet]

Hi Phil,

I had tried

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"${PLACE}"
but I had not tried

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'"${PLACE}"'

Since the whole

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'-filename< ... '
Was in single quotes, I assumed that having single quotes in the middle would break something.  That is what I get for assuming.

Your example worked.  Thanks!

[Phil Harvey]

All I did was change quotes from single to double then back again.

'abc'"abc"'abc' is just the concatenation of three strings with different quoting.

- Phil