PHP - Json_encode 如何将变量添加到数据库选择并传递给 Json？

Question

我需要从 MySQL 数据库中获取数据，在其附近添加一个变量，然后以 JSON 格式显示。

输出应如下所示：

{"items":[{"cinemaname":"Cinema name","logo":"upload/cinemaname.png","distance":"103"},{"cinemaname":"Cinema name 2","logo":"upload/cinemaname2.png","distance":"23"}]}

表格不包含距离，需要在代码中手动计算并添加距离。

代码：

$result = mysqli_query($con,"some query here");

while($rowm = mysqli_fetch_array($result))
{
    $all[]= $rowm;
}

$alldata = array('items'=>$all);
header('Content-Type: application/json; charset=utf-8');
echo json_encode($alldata);
}

如何将距离添加到输出中？

我试过这个：

$result = mysqli_query($con,"some query here");

while($rowm = mysqli_fetch_array($result))
{
    //here is calculation of a distance = $distance

    $all[]= $rowm.'distance=>'.$distance;    
}

$alldata = array('items'=>$all);
header('Content-Type: application/json; charset=utf-8');
echo json_encode($alldata);
}

目前，脚本输出如下：

{"items":[{"cinemaname":"Cinema name","logo":"upload/cinemaname.png"},{"cinemaname":"Cinema name 2","logo":"upload/cinemaname2.png"}]}

我希望它看起来像这样：

{"items":[{"cinemaname":"Cinema name","logo":"upload/cinemaname.png","distance":"103"},{"cinemaname":"Cinema name 2","logo":"upload/cinemaname2.png","distance":"23"}]}

Answer 1

$rowm 是一个数组，因此通过. 使用字符串连接将不起作用！请试试这个：

$rowm['distance'] = $distance;
$all[] = $rowm;

希望对你有帮助。

Answer 2

$result = mysqli_query($con,"SELECT * FROM `items`");

while($rowm = mysqli_fetch_assoc($result))
{
    $all['items'][] = array_values($rowm);
}

header('Content-Type: application/json; charset=utf-8');
echo json_encode($all);
die();

Answer 3

$result = mysqli_query($con,"some query here");

while($rowm = mysqli_fetch_array($result))
{
    $distance={calc of distance};
    $all[]=array(
            "cinemaname"=>$rowm['name']
        ,   "logo"=>$rowm['logo']
        ,   "distance"=>$distance); 
}
$alldata = array('items'=>$all);
header('Content-Type: application/json; charset=utf-8');
echo json_encode($alldata);