Test tables:
(what if you would have to set the records, as you comment the tables have data, if you want to make id_tariff is fk and in general you have a id_tarifa but in the tariff table that id does not exist is going to give an error because it will not find the key foreigner)
use prueba;
create table tarifa(
id_tarifa int not null auto_increment,
proveedor nvarchar(50) not null,
desc_tar nvarchar(50) not null,
primary key (id_tarifa)
);
create table general(
id_general int not null auto_increment,
nombre nvarchar(50) not null,
id_tarifa int not null,
primary key (id_general),
FOREIGN KEY (id_tarifa)
REFERENCES tarifa (id_tarifa)
);
Now creating it without FK and using the line you pass you also works:
use prueba;
create table tarifa(
id_tarifa int not null auto_increment,
proveedor nvarchar(50) not null,
desc_tar nvarchar(50) not null,
primary key (id_tarifa)
);
create table general(
id_general int not null auto_increment,
nombre nvarchar(50) not null,
id_tarifa int not null,
primary key (id_general)
);
ALTER TABLE general ADD FOREIGN KEY (id_tarifa) REFERENCES
tarifa(id_tarifa)
Now if the tables have data to be able to create the FK you need all the data, look at the following example, I create the table without FK charge two data in general and one only in tariff, one of the data in general refers to rate 2 but this does not exist in the tariff table, now I execute the command to do the FK and there I get an error I do not give me because it does not find all the references
Try this query (fit the exact names of your tables)
SELECT G.id_general, T.id_tarifa from GENERAL G LEFT JOIN TARIFA T ON
T.id_tarifa = g.id_tarifa
In my example it shows me that there are two registers but for register two there is no fee two, there is the error in my example
