Access column returned by psql from bash

Question

I want to use value returned by psql in bash. This is the script:

psql "connection parameters" -c "SELECT pg_database_size('dbname');"

The output is like this:

 pg_database_size 
------------------
          5773072
(1 row)

But I only want the 5773072 so I can use it in logging. Can anyone help?

Milen A. Radev · Accepted Answer · 2012-03-11 12:57:38Z

3

The -t (--tuples-only) option might be used also:

psql "connection parameters" -t -c "SELECT pg_database_size('dbname');"

answered Mar 11, 2012 at 12:57

Milen A. Radev

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Erwin Brandstetter · Accepted Answer · 2022-05-11 23:21:33Z

5

In addition to what @Milen already provided, you may want -A (--no-align) to remove leading white space:

psql "con params" -tAc "SELECT pg_size_pretty(pg_database_size('mydb'))"

As an aside: pg_size_pretty() may or may not be of interest.

answered Mar 12, 2012 at 6:32

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1

I think this is better than the confirmed answer
– RustyShackleford
Feb 13, 2018 at 21:00

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Majid Azimi · Accepted Answer · 2012-03-11 12:28:00Z

1

I got it:

psql "connection parameters" -c "SELECT pg_database_size('dbname');" | tail -3 | head -1 | tr -d ' '

answered Mar 11, 2012 at 12:28

Majid Azimi

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3 Answers 3