PHP - 无法访问数组值

Question

我正在尝试使用 PHP 访问 MySQL 数据库，但遇到数据问题。在我的开发机器(PHP 版本 5.5)上，这段代码运行得很好，但在生产服务器(PHP 版本 5.6)上，代码在处理数组时似乎失败了。基本上，我似乎得到下面的代码来 echo 表明 $parks 确实是一个数组，并添加

foreach ($parks as $key => $val) {
    echo $key;
};

将正确回显数组键，但如果我将 echo $key 替换为 echo $val，则什么也不会发生，根本没有回显。

<?php
    $dbname = "DATABASE";
    $password = "PASSWORD";
    $servername = "localhost";
    $dbname = "NAME";

    $conn = new mysqli($servername, $username, $password, $dbname);

    $query = "SELECT name, latitude, longitude, description, where_to_go from locations";
    $result = mysqli_query($conn, $query);

    $parks = [];

    while($row = mysqli_fetch_array($result)){                   
        $park = new stdClass();
        $location = new stdClass();
        $location->lat = floatval($row['latitude']);
        $location->lng = floatval($row['longitude']);
        $park->title = $row['name'];
        $park->location = $location;
        $park->description = $row['description'];
        $park->whereToGo = $row['where_to_go'];
        array_push($parks, json_encode($park));
    };
    /*
    foreach ($parks as $key => $val) {
        echo $key; //THIS WILL WORK
    };

    foreach ($parks as $key => $val) {
        echo $val; //THIS WILL NOT RETURN ANYTHING
    };

    echo $parks[0] //SIMILARLY, THIS WILL NOT RETURN ANYTHING
    */

    $conn->close();
?>

为了进一步说明这一点，我的程序的下一步是将 $parks 数组放入 JavaScript 变量中...

<script type="text/javascript">
    var locations = <?php echo json_encode($parks); ?>;
    var locations = locations.map(JSON.parse);
    console.log(locations);
</script>

运行所有这些都会记录[false]。

为了测试，我在 PHP 的 while 循环中针对 $park 的不同属性添加了一些 echo ，它们输出正确的值...所以数组似乎构建正确，但我似乎无法从中得到任何东西。

非常感谢任何指导。

Answer 1

请尝试一下:

while($row = mysqli_fetch_array($result)){                   
    $park = new stdClass();
    $location = new stdClass();
    $location->lat = floatval($row['latitude']);
    $location->lng = floatval($row['longitude']);
    $park->title = $row['name'];
    $park->location = $location;
    $park->description = $row['description'];
    $park->whereToGo = $row['where_to_go'];
    $parks[] = json_encode(get_object_vars($park));
};

或者只使用数组:

while($row = mysqli_fetch_array($result)){                   
    $park = [];
    $location = [];
    $location['lat'] = floatval($row['latitude']);
    $location['lng'] = floatval($row['longitude']);
    $park['title'] = $row['name'];
    $park['location'] = $location;
    $park['description'] = $row['description'];
    $park['whereToGo'] = $row['where_to_go'];
    $parks[] = json_encode($park);
};