0
Code to insert:
$codigoutente = $_POST['codigoutente'];
$codvalencia = $_POST['codvalencia'];
$Responsavel = $_POST['Responsavel'];
$Contato = $_POST['Contato'];
for ($i=0;$i<count($_POST["Responsavel"]);$i++) {
$Responsavel = $_POST['Responsavel'][$i];
$stmt1 = $conn->prepare("INSERT INTO Responsaveis (`Responsavel`,`IdUtente`) VALUES ('$Responsavel','$codigoutente')");
mysqli_stmt_execute($stmt1);
}
$last_id = $conn->insert_id;
for ($i=0;$i<count($_POST["Contato"]);$i++) {
$Contato = $_POST['Contato'][$i];
$stmt2 = $conn->prepare("INSERT INTO ContatoRes (`IdUtente`,`IdResponsavel`,`Contato`) VALUES ('$codigoutente','$last_id','351$Contato')");
mysqli_stmt_execute($stmt2);
}
The problem is the variable $last_id
. When I insert two new guardians into the first insert
will insert two lines with two ids
different, example:
- id = 10 Idresponsable = Alberto Idutente = 10115
- id = 11 Idrsponsable = Arming Idutente = 10118
Then when doing the second insert
the id taken from the previous table is always in the latter for the two records and cannot, you have to take the first insert
the id = 10
and in the second the id = 11
.
$_POST["Contato"]
and$_POST["Responsavel"]
are arrays that are receiving ?– Bulfaitelo