Concatenate JSON in php with read Database - php

Question

I read user from the database but I return json to ajax only return the last user because I cant concatenate the json encode.
$myObj = new \stdClass();
while ($fila = $bd->fila()) {
$myObj->NOMBRE = $fila["NOMBRE"];
$myObj->ROLENAME = $fila["ROLENAME"];
$myObj->IDUSER = $fila["IDUSER"];
$myJSON = json_encode($myObj);
}
echo $myJSON;

Answer 1

You are now overwriting $myJson in each iteration. We can also not "concatanate" two jsons (well, we could, but we shouldn't, cause it's laborious..).
Better put everything in one array/object and json_encode() at the very end.
$myObj = new \stdClass();
$users = array(); // instantiate a new array
while ($fila = $bd->fila()) {
$myObj->NOMBRE = $fila["NOMBRE"];
$myObj->ROLENAME = $fila["ROLENAME"];
$myObj->IDUSER = $fila["IDUSER"];
// add objects to that array
$users[] = $myObj;
}
// then at the end encode the whole thing to json:
echo json_encode($users);
Now here are some variants:
If you want everything that your db is returning in this items you could shorten that to:
$users = array(); // instantiate a new array
while ($fila = $bd->fila()) {
// add the whole item (cast as Object) to that array
$users[] = (Object) $fila;
}
// then at the end encode the whole thing to json:
echo json_encode($users);
If you don't care if the items are objects or arrays you could skip the casting and just add the array to the big array:
$users[] = $fila;

Answer 2

Maybe if you concatenate it as array like this
$myJSON[] = $myObj;
and then after the while
echo json_encode($myJSON);

Answer 3

You just have to collect data to big array and then encode whole array. Also there's no reason for creating new StdObjects instead of usual arrays.
// define empty array
$result = [];
while ($fila = $bd->fila()) {
// add to the array all needed elements one by one
$result[] = [
'NOMBRE' => $fila["NOMBRE"],
'ROLENAME' => $fila["ROLENAME"],
'IDUSER' => $fila["IDUSER"],
];
}
// encode whole result
echo json_encode($result);