Question

这是我在网上的mysql数据库

这是访问数据库以获取JSON文件的PHP文件，首先我选择表中的所有“类别”单元格，然后遍历该类别，然后遍历该类别的其他列，然后进行分配到$ data2数组，然后将$ data2和category分配给br一个$ data数组，这是我要显示的json。

<?php 
header("Access-Control-Allow-Origin: *");
$user = "u348833238_rest"; /* User */
$password = "a!23286029"; /* Password */
$dbname = "u348833238_rest"; /* Database name */
$host = "localhost";

$con = mysqli_connect($host, $user, $password, $dbname);
// Check connection
if (!$con) {
 die("Connection failed: " . mysqli_connect_error());
} 

// $sel = mysqli_query($con,"select * from restaurant");

// $data = array();

// while ($row = mysqli_fetch_array($sel)) {
//  $data[] = array("dishes" => ["name"=>$row['food'], "price"=>$row['price']] , "category"=>$row['category']);
// }
// echo json_encode($data);


$sel = mysqli_query($con,"select distinct category from restaurant");

$data = array();

while ($row = mysqli_fetch_array($sel)) {

    $c = $row['category'];

    $sel2 = mysqli_query($con,"select * from restaurant where category = $c ");

    $data2 = array();

    while ($row2 = mysqli_fetch_array($sel2)){
        $data2[] = array("name"=>$row2['food'], "price"=>$row2['price']);
    }

    // echo $data2;

    $data[] = array("category"=>$row['category'], "dishes"=>$data2);
}

// echo $data;
echo json_encode($data);

?>

由于属性“ dishes”的数组为空，JSON如何无法正确显示，如下所示：

the array is empty

Answer 1

我建议您打开错误，以便您查看问题出在哪里。

猜测在这里：

select * from restaurant where category = $c

应该是：

select * from restaurant where category = '$c'

来自MySQL的JSON无法通过PHP正确显示

1 个答案: