Variable substitution is not working

[uk_jih]

I have the following code:

for (( i=5; i>=0; i=i-1 ))
do
sqlfile=`date +%Y-%m-%d --date='$i days ago'`
echo $sqlfile
done

How can I get the $i replaced with the contents of $i, everything I try seems to fail

Cheers
John

[lomcevak]

Use double quotes instead of single quotes.

sqlfile=`date +%Y-%m-%d --date="$i days ago"`

Single quotes prevent variable replacement.

[uk_jih]

Thanks for that.

I just assumed that the single quotes had to be there for the command to work.

It worked with double quotes too.

John