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Accepted
LukeR
Asked: 2011-09-08 19:58:52 +0800 CST

Bash variable assignment using another variable as part

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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.

scripting bash

3 Answers

  1. Best Answer

    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.

    • 8

  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
    • 4

  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
    • 2