无法从数据库中检索输出数据

Question

我使用以下PHP代码从数据库中检索商店信息字段，但是当我运行PHP时，我得到了如下所示的错误。

<?php
include ('database.php');


$locLat ;
$locLng ;
$shopName;
$shopContact;

$sql = "SELECT s_iD, s_name,s_contNo,s_lat,s_long from tbl_shop";

$result = mysqli_query($con,$sql);
$arrayResult = array();

while ($row = mysqli_fetch_array($result)){
    array_push($arrayResult,array("id"=>$row['s_iD'],"shopName"=>$row['s_name'],"shopContact"=>$row['s_contNo'],"latitude"=>$row['s_lat'],
        "longitude"=>$row['s_long']));
}

echo json_encode (array('result'=> $result));


mysqli_close($con);

但输出总是：

  

{ “结果”：{ “current_field”：NULL， “场计数”：NULL， “长度”：NULL， “NUM_ROWS”：空， “类型”：空}}

Answer 1

这里更新的代码应该是：

<?php
include ('database.php');


$locLat ;
$locLng ;
$shopName;
$shopContact;

$sql = "SELECT s_iD, s_name,s_contNo,s_lat,s_long from tbl_shop";

$result = mysqli_query($con,$sql);
$arrayResult = array();

while ($row = mysqli_fetch_array($result)){
    array_push($arrayResult,array("id"=>$row['s_iD'],"shopName"=>$row['s_name'],"shopContact"=>$row['s_contNo'],"latitude"=>$row['s_lat'],
        "longitude"=>$row['s_long']));
}

echo json_encode (array('result'=> $arrayResult));  // <-- changed from $result


mysqli_close($con);