Bash variable assignment using another variable as part

Question

I have the following excerpt from a bash script used to backup a database

#!/bin/bash DB=database_name DBUSER=username FILENAME=$DB_$(date +%s).sql 

I am trying to reuse the value of DB within the FILENAME variable assignment, but it won't let me use substitution this way. I just get the timestamp for the file name.

Is it possible to achieve what I want, and if so what is the syntax?

thanks.

Answer 1

The problem is that bash does not know that you mean $DB instead of $DB_ (which is a perfectly valid name for a variable).

The best option is to be explicit on the name of the variable using braces around its name:

FILENAME=${DB}_$(date %s).sql 

This saves you the trouble of escaping other characters which are not to be interpreted as part of a variable name.

Answer 2

Alternately, use braces to isolate your variable. I think it's a bit clearer than putting quotes in there.

FILENAME=${DB}_$(date +%s).sql 

Answer 3

Enclose the underscore in double quotes (or single quotes for the stronger):

#!/bin/bash DB=database_name DBUSER=username FILENAME=$DB"_"$(date +%s).sql