JSON：查询连接表不起作用（Cordova Mobile）

Question

I have 2 query the result of which I will use in JSON . The first query is

$query = "SELECT * 
            FROM info_lokasi 
           WHERE id_lokasi=".$id."";

and the second query

$query = "SELECT t1.id_lokasi,t1.jalan,t1.lat,t1.lng,t1.nama,t2.kapasitas,t2.kegiatan,t2.parkir,t2.pengurus
            FROM info_lokasi t1 
      INNER JOIN detail t2 ON t1.id_lokasi=t2.id_lokasi
           WHERE t1.id_lokasi 
             AND t2.id_lokasi=".$id."";

when I use the first query , the result I get is

[
{
"id_lokasi": "3",
"nama": "masjid",
"jalan": "surabaya",
"lat": "-7.292958471571183",
"lng": "112.71161556243896",
"username": "ope",
"jenis": "masijd"
}
]

but, when I use a second query , I did not get the result. WHY?

this is my code:

PHP

<?php 
    include "db.php";
    header("Content-type: text/javascript");

    if (isset($_POST['detail'])) {
    $id = $_POST['id'];

    $query = "SELECT t1.id_lokasi,t1.jalan,t1.lat,t1.lng,t1.nama,t2.kapasitas,t2.kegiatan,t2.parkir,t2.pengurus FROM info_lokasi t1 INNER JOIN detail t2 ON t1.id_lokasi=t2.id_lokasi WHERE t1.id_lokasi AND t2.id_lokasi=".$id."";

    //$query = "SELECT * FROM info_lokasi WHERE id_lokasi=".$id."";

    $data = $con->query($query);

    $results = array();
    while ($r = mysqli_fetch_array($data,MYSQL_ASSOC)) {
        # code...
        $results[] = $r;
    }
    echo json_encode($results);
    }
    mysqli_close($con);

?>

Javascript

function detail(idlokasi){
    var id = idlokasi;
    var datastring = "id="+id+"&detail=";
            $.ajax({
                type: "POST",
                url: "http://localhost:8080/maps-api/detail.php",
                data: datastring,
                crossDomain: true,
                cache: false,
                datatype: "JSON",
                success: function(data){
                    alert("DATA: "+data);                
                } 
            });
}

Thanks

Answer 1

The reason why you don't get any output can be:

1. Query really returns no data (i.e.: no rows match WHERE conditions)
2. There's an error on your SQL

As you expected it to return data I'll assume that is case 2.

Your query is:

      SELECT t1.id_lokasi,t1.jalan,t1.lat,t1.lng,t1.nama,t2.kapasitas,t2.kegiatan,t2.parkir,t2.pengurus
        FROM info_lokasi t1 
  INNER JOIN detail t2 ON t1.id_lokasi=t2.id_lokasi
       WHERE t1.id_lokasi 
         AND t2.id_lokasi=$id

To detect the error, you should use this PHP code:

$query = "SELECT t1.id_lokasi,t1.jalan,t1.lat,t1.lng,t1.nama,t2.kapasitas,t2.kegiatan,t2.parkir,t2.pengurus
        FROM info_lokasi t1 
  INNER JOIN detail t2 ON t1.id_lokasi=t2.id_lokasi
       WHERE t1.id_lokasi 
         AND t2.id_lokasi=$id";
$data = $con->query($query);
if ($con->error)
{
    echo "ERROR: {$con->error}";
    throw new Exception($con->error);
}

Note: don't use $_POST['id'] as your parameter as it can lead to SQL injection problems. You should better use query parameters.