How to count boolean values in PostgreSQL?

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  •  30-12-2020
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Question

How to count, how may true and false for the field public in postgresql user table i have tried this query

select 
sum(case when false then 1 else 0 end) as false, 
sum(case when true then 1 else 0 end) as true
from  public.user;

but am not getting any value and if i remove public from query then i will get correct counts only i have value true 

table : name|   DOB     | public
values : bb | 20/2/1991/| true

op : true = 1 and false  = 0

but am getting the same answer when i make public as false 

table : name|   DOB     | public
values : bb | 20/2/1991/| false

op : true = 1 and false  = 0

so someone please help me to solve this

Was it helpful?

Solution

Use the filter() clause:

select count(*) filter (where "public") as public_count,
       count(*) filter (where not "public") as not_public_count
from  public."user";

Note that user is a reserved keyword, you have to use double quotes in order to use it as a table name.

The above assumes that the column public is of type boolean

OTHER TIPS

Other ways, that work in older versions that don't have FILTER, using CASE expressions or subqueries:

SUM and CASE expression

select
    sum(case when not public then 1 else 0 end) as false,
    sum(case when     public then 1 else 0 end) as true
from 
    public.user;

COUNT and CASE expression (the default ELSE NULL is omitted)

select
    count(case when not public then 1 end) as false,
    count(case when     public then 1 end) as true
from
    public.user;

SUM after converting the boolean to integer (TRUE -> 1, FALSE -> 0)

select
    sum((not public)::int) as false,
    sum(     public ::int) as true
from
    public.user;

a rather obfuscated solution (using 3VL to convert FALSE to NULL)
(by @Andriy):

select
    count(not public or null) as false,
    count(    public or null) as true
from
    public.user;

a slightly more clear (or more obfuscated?) 3VL abuse:

select
    count(public and null) as false,
    count(public or  null) as true
from
    public.user;

a subquery for each count

select
    (select count(*) from public.user where not public) as false,
    (select count(*) from public.user where     public) as true
 ;

Case expression has two possible forms:

case <field> when <value> then <actions1> else <actions2> end
case when <condition> then <actions1> else <actions2> end

In this clause, the condition is never true

case when false then 1 else 0 end

Instead, use

case public."user".public when false then 1 else 0 end

or

case when public."user".public = false then 1 else 0 end

You don't get proper results because you omitted the column from your expressions:

select 
sum(case when "public" = false then 1 else 0 end) AS false_ct, 
sum(case when "public" = true  then 1 else 0 end) AS true_ct
from  public.user;

There are various simpler and faster techniques. Suggestions in other answers. It's a recurring topic ...

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