无法从数据库检索输出数据

Cross 发表于 Dev

叉

我正在使用以下 PHP 代码从数据库中检索商店信息字段，但是当我运行 PHP 时出现如下所示的错误。

<?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：

{"result":{"current_field":null,"field_count":null,"lengths":null,"num_rows":null,"type":null}}

巴拉特帕尔玛

这是您更新的代码应该是：

<?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);

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