Passing variable to array

Question

I am passing data base objects to an array.

I need to include another variable to the array. The variable is $latitud_usuario.

Here is the code:

if ($result->num_rows > 0) {
        while ($obj = $result->fetch_object()) {
            $arr[] = array('nombre_doctor' => $obj->nombre_doctor,'apellido1_doctor' => $obj->apellido1_doctor,'apellido2_doctor' => $obj->apellido2_doctor,'ciudad_doctor' => $obj->ciudad_doctor, 'latitud_doctor' => $latitud_usuario);
        }
    }
}
echo json_encode($arr);

If I create the array including only fetched objects, the JSON sent is ok, but after including the last array object:

'latitud_doctor' => $latitud_usuario

the JSON is not received as it should.

I guess this last array object expression is wrong.

Any hint is welcome.

can you show the json_encode($arr); output?and is there any errors? — Madhawa Priyashantha
– Madhawa Priyashantha, Commented May 2, 2016 at 4:53
please print_r array before encoding, at least the last portion, from where $latitud_usuario come from and what is the content in it? — Murad Hasan
– Murad Hasan, Commented May 2, 2016 at 4:55
i guess it is $obj -> latitud_usuario (instead of $latitud_usuario) — Mohammedshafeek C S
– Mohammedshafeek C S, Commented May 2, 2016 at 4:56
@FastSnail, it is ok, the problem lies at the value of $latitud_usuario, not at the array...thanks — mvasco
– mvasco, Commented May 2, 2016 at 5:00
@mvasco enable errors.so if there is a problem you can see it — Madhawa Priyashantha
– Madhawa Priyashantha, Commented May 2, 2016 at 5:05

sujivasagam · Accepted Answer · 2016-05-02 04:59:21Z

1

Try this

if ($result->num_rows > 0) {
        while ($obj = $result->fetch_object()) {
            $arr[] = array('nombre_doctor' => $obj->nombre_doctor,'apellido1_doctor' => $obj->apellido1_doctor,'apellido2_doctor' => $obj->apellido2_doctor,'ciudad_doctor' => $obj->ciudad_doctor, 'latitud_doctor' => $latitud_usuario);
           $arr['latitud_doctor']=$latitud_usuario;
        }
    }
}
echo json_encode($arr);

answered May 2, 2016 at 4:59

sujivasagam

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2 Comments

Murad Hasan Over a year ago

don't think this is good answer. in your array you have 'latitud_doctor' => $latitud_usuario and outer the array add the $arr['latitud_doctor']=$latitud_usuario;.

Madhawa Priyashantha Over a year ago

can you tell why 'latitud_doctor' => $latitud_usuario doesn't work ? it should work.

Webomatik · Accepted Answer · 2016-05-02 05:09:20Z

0

Here's a version that works (using a dummy $obj object):

$obj = (object) array('nombre_doctor'=> 6, 'apellido1_doctor' => 'whatever1', 'apellido2_doctor'  => 'whatever2', 
'ciudad_doctor' => 'Montreal', 'latitud_usuario' => '35463');
$arr[] = array('nombre_doctor' => $obj->nombre_doctor,'apellido1_doctor' => $obj->apellido1_doctor,
'apellido2_doctor' => $obj->apellido2_doctor,'ciudad_doctor' => $obj->ciudad_doctor, 
'latitud_doctor' => $obj->latitud_usuario);

echo json_encode($arr);

answered May 2, 2016 at 5:09

Webomatik

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