Trying to store information in a database with two tables, but relating the tables with a foreign key, but I have not been able to insert the information.
The code is as follows:
insertarTablas: function(db) {
var datosPersona, datosTelefono, insertarPersona, insertarTelefono, largoPer, largoTel, i, j;
datosPersona = [{
cedula: 123456,
nombre: "pepe",
apellido: "perez"
}];
datosTelefono = [{
numero: 30020025,
cedula: 123456
}, {
numero: 6998877,
cedula: 123456
}, {
numero: 58965475,
cedula: 123456
}];
insertarPersona = "INSERT INTO Persona(cedula, nombre, apellido) VALUES(?,?,?)";
insertarTelefono = "INSERT INTO Telefono(numero, cedula_tel) VALUES(?,?)";
largoPer = datosPersona.length;
largoTel = datosTelefono.length;
for (i = 0; i < largoPer; i++) {
$cordovaSQLite.execute(db, insertarPersona, [datosPersona[i].cedula, datosPersona[i].nombre, datosPersona[i].apellido]);
}
for (i = 0; i < largoTel; i++) {
$cordovaSQLite.execute(db, insertarTelefono, [datosTelefono[i].numero, datosTelefono[i].cedula]);
}
}
In the following image, point out the results of operations with rectangles

The red color is the record that is inserted of the person, bone the person object.
The green color, shows that, it shows that inserted in three the three register, always starting from position 0.
The sky blue, the number of rows affected by the query, knowing that I put as a criterion * that specifies all, and therefore I throw 1, which is the number of records.
Lastly a purple color, which shows a single record in the Telephone tab, which is supposed to hold 3.
For help please post a complete, self contained test program that demonstrates your issue.
@brodybits sorry, i solved.
Thanks!
Hi @PterPmnta,
I have used following lines of code for create two tables with foreign key,
CREATE TABLE artist(artistid INTEGER PRIMARY KEY, artistname TEXT);
CREATE TABLE IF NOT EXISTS track(trackid INTEGER PRIMARY KEY,trackname TEXT,trackartist INTEGER,FOREIGN KEY(trackartist) REFERENCES artist(artistid));
But foreign key(artistid) not stored in track table.
I has try the following code to enable.
query('PRAGMA foreign_keys = ON;');
But no effect, could you please help me to solve this.