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 数据库获取数据，在其附近添加一个变量，然后以 JSON 格式显示它。

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"}]}

该表不包含距离，距离应在代码中手动计算并添加。

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!请尝试这个:

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

希望这对您有帮助。