How do I count the same results and add the amount up front?
For example:
Currently the records in my table look like this:
Id | Nome
------------
0 | Teste
1 | Teste
2 | Teste
I want to give DISTINCT to them and leave them like this:
Id | Nome
------------
0 | Teste(3)
SQLFIDDLE for testing
I am using MySql !
You can try to use the following sql expression:
select distinct id, CONCAT(nome, "(",count(id),")") as Nome from teste group by nome;
I looked for the same help and this was a solution that worked.
SELECT DISTINCT campo1
,campo2
,campo3
,campo4
,campo5
,campo6
,campo7
,COUNT(*) AS quantidade
FROM tabela_a
GROUP BY campo1
,campo2
,campo3
,campo4
,campo5
,campo6
,campo7
ORDER BY quantidade DESC
Credits: Luan Moreno [SQL Soul] || SQL Server Developer || MCTS SQL Server Admin and Dev @luansql
See if this helps you:
select COUNT(distinct Nome) from @TableTeste
In SQLFIDDLE Test:
create table algo(
id int not null auto_increment,
nome varchar(10) not null,
primary key(id));
insert into algo (id, nome) values (0,'Teste');
insert into algo (nome) values (1,'Teste');
insert into algo (nome) values (2,'Teste');
select distinct concat(nome,'(',(select count(id) from algo),')') as nome from algo;
select distinct concat(nome,'(',count(id),')') as nome from algo;
select distinct sum(id-id) as id, concat(nome,'(',count(id),')') as nome from algo group by nome;