无法检索数据库中的输出数据

Question

我正在使用以下PHP代码从数据库检索商店信息字段，但在运行PHP时遇到下面显示的错误。

<?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); 

但是输出总是：

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

Answer 1

这里是你更新的代码应该是：

<?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);