Check if database exists in postgreSQL using shell

I use the following modification of Arturo's solution:

psql -lqt | cut -d \| -f 1 | grep -w <db_name> | wc -l

What it does

psql -l outputs something like the following:

                                        List of databases
     Name  |   Owner   | Encoding |  Collate   |   Ctype    |   Access privileges   
-----------+-----------+----------+------------+------------+-----------------------
 my_db     | my_user   | UTF8     | en_US.UTF8 | en_US.UTF8 | 
 postgres  | postgres  | LATIN1   | en_US      | en_US      | 
 template0 | postgres  | LATIN1   | en_US      | en_US      | =c/postgres          +
           |           |          |            |            | postgres=CTc/postgres
 template1 | postgres  | LATIN1   | en_US      | en_US      | =c/postgres          +
           |           |          |            |            | postgres=CTc/postgres
(4 rows)

Using the naive approach means that searching for a database called "List, "Access" or "rows" will succeed. So we pipe this output through a bunch of built-in command line tools to only search in the first column.

The -t flag removes headers and footers:

 my_db     | my_user   | UTF8     | en_US.UTF8 | en_US.UTF8 | 
 postgres  | postgres  | LATIN1   | en_US      | en_US      | 
 template0 | postgres  | LATIN1   | en_US      | en_US      | =c/postgres          +
           |           |          |            |            | postgres=CTc/postgres
 template1 | postgres  | LATIN1   | en_US      | en_US      | =c/postgres          +
           |           |          |            |            | postgres=CTc/postgres

The next bit, cut -d \| -f 1 splits the output by the vertical pipe | character (escaped from the shell with a backslash), and selects field 1. This leaves:

 my_db             
 postgres          
 template0         

 template1         

grep -w matches whole words, and so won't match if you are searching for temp in this scenario. And the wc -l counts matchine lines, yielding 1 if the db exists, and 0 if it doesn't.

If you'd rather follow the UNIX convention of returning an exit code rather than printing output, drop the wc from the pipeline:

psql -lqt | cut -d \| -f 1 | grep -w <db_name>

The exit status of that pipeline will be 0 (success) if the database exists or 1 (failure) if it doesn't. Your shell will set the special variable $? to the exit status of the last command. You can also test the status directly in a conditional:

if psql -lqt | cut -d \| -f 1 | grep -w <db_name>; then
    # database exists
    # $? is 0
else
    # ruh-roh
    # $? is 1
fi

The following shell code seems to work for me:

if [[ `psql -tAc "SELECT 1 FROM pg_database WHERE datname='DB_NAME'"` == "1" ]]
then
    echo "Database already exists"
else
    echo "Database does not exist"
fi

I'm new to postgresql, but the following command is what I used to check if a database exists

if psql ${DB_NAME} -c '\q' 2>&1; then
   echo "database ${DB_NAME} exists"
fi

#!/bin/sh
DB_NAME=hahahahahahaha
psql -U postgres ${DB_NAME} --command="SELECT version();" >/dev/null 2>&1
RESULT=$?
echo DATABASE=${DB_NAME} RESULT=${RESULT}
#

I'm still pretty inexperienced with shell programming, so if this is really wrong for some reason, vote me down, but don't be too alarmed.

Building from kibibu's answer:

# If resulting string is not zero-length (not empty) then...
if [[ ! -z `psql -lqt | cut -d \| -f 1 | grep -w $DB_NAME` ]]; then
  echo "Database $DB_NAME exists."
else
  echo "No existing databases are named $DB_NAME."
fi

For completeness, another version using regex rather than string cutting:

psql -l | grep '^ exact_dbname\b'

So for instance:

if psql -l | grep '^ mydatabase\b' > /dev/null ; then
  echo "Database exists already."
  exit
fi