-1
I have the following SQL that correctly counts the number of properties:
select
clientes.id,
clientes.nome,
clientes.status,
clientes.cliente,
clientes.tipo,
clientes.disponibilidade,
imoveis.id,
imoveis.cod,
imoveis.status,
imoveis.vvenda,
COUNT(imoveis.id) AS imoveis
from clientes
inner join imoveis on clientes.cliente = imoveis.cod
where
imoveis.status='2'
AND clientes.status='2'
AND imoveis.vvenda < clientes.disponibilidade
AND imoveis.vvenda <> '0'
AND clientes.cliente = '$cliente'
AND imoveis.cod = '$cliente'
GROUP BY clientes.id
Only that I had to make a small implementation (I highlighted below what I added) only that the number of properties is now incorrect in the count. See below the SQL with implementations:
select
clientes.id,
clientes.nome,
clientes.status,
clientes.cliente,
clientes.tipo,
clientes.disponibilidade,
imoveis.id,
imoveis.cod,
imoveis.status,
imoveis.vvenda,
COUNT(imoveis.id) AS imoveis,
photos.Cod
from clientes
inner join imoveis on clientes.cliente = imoveis.cod
Inner Join photos on photos.Cod=immobles.id
where
imoveis.status='2'
AND clientes.status='2'
AND imoveis.vvenda < clientes.disponibilidade
AND imoveis.vvenda <> '0'
AND clientes.cliente = '$cliente'
AND imoveis.cod = '$cliente'
GROUP BY clientes.id
These simple implementations were enough for the property count to be totally different.
I really need help!
The value is different as? Higher or lower than the value "correct", what is the relationship between photo and immovable? 1 for 1, 1 for N?
– Rubico
What is the criterion that the personnel is using to negatively? The person who negatively could inform their motives here?
– Rubico
@Rubico The value is different in the quantity of filtered properties. In the first example COUNT(immovables.id) Immovable AS displays perfectly the number of properties. In the second example, just because I made the small implentations the COUNT(immovel.id) Immovables displays totally different values, ie incorrect.
– Gladison Neuza Perosini
without a knowledge of your bank is a little difficult, but I noticed that you are comparing the
foto.cod
with theimovel.id
, would not be with theimovel.cod
? Try to show what is the structure of your bank, what are the relationships.– Rubico
What you have valved is now valid. As there is a relationship between photo properties (where a property can have several photos), it is necessary that, in the table of photos, there is a column to store which property is the "owner" of the photo, that is, in the table of photos there must be a column
fotos.imovel_cod
, for example, and it should be through this column that theinner join
be done (INNER JOIN fotos ON fotos.imovel_cod = imovel.cod
). How is the structure of the table photos? There is this column to make the relationship between photos and real estate?– Márcio Lordelo