Question

我正在使用Codeigniter作为我的框架，我已经查看了文档，但我找不到解决方案。

现在我已经从数据库中选择了数据，我需要在检索到的每条记录的末尾显示距离。

Example of json: {"details":[{"country":"United States","_id":"3892","lat":"39.954559","lng":"-82.837608","admin_level_1":"Ohio","locale":"Columbus"}]}

这就是它需要的东西，请记住，距离不在数据库中，而是在运行中计算。

Example of json: {"details":[{"country":"United States","_id":"3892","lat":"39.954559","lng":"-82.837608","admin_level_1":"Ohio","locale":"Columbus", "distance": "1.2 Mi"}]}

有关如何将距离的任何想法附加到每个结果的末尾？

        $dbStations->select('lat');
        $dbStations->select('lng');
        $dbStations->select('_id');
        $stations = $dbStations->get('stDetails');
        foreach ($stations->result() as $station) {
            $lat2 = $station->lat;
            $lng2 = $station->lng;
            $stId = $station->_id;
            $distance = $this->distance->gpsdistance($lat1, $lng1, $lat2, $lng2, "M");
                if ($distance <= $rad) {
                //at this point the code has determined that this id is within
                //the preferred distance.
                $stationArr[] = $stId;
            }
        }
            $dbStations->select('country, _id, lat, lng, admin_level_1, locale');
            $dbStations->where_in('_id', $stationArr);
            $stations = $dbStations->get('stationDetails');
        echo json_encode(array('stations' => $stations->result()));
    }

Answer 1

您可以尝试一次完成所有事情。根据可能知道您的表格，此查询应该为您提供半径$rad内的工作站。

必须定义变量$lat1和$lng1。

$query= "select country, _id, lat, lng, admin_level_1, locale, 
    (acos(sin(radians($lat1)) * sin(radians(lat)) + cos(radians($lat1)) *
    cos(radians(lat)) * cos(radians($lng1) - radians(lng))) * 6378)
     as distance
    from stationDetails
    where (acos(sin(radians($lat1)) * sin(radians(lat)) + 
          cos(radians($lat1)) * cos(radians(lat)) * cos(radians($lng1) - 
          radians(lng))) * 6378)<=$rad
    order by distance desc";

echo json_encode($this->db->query($query)->result_array());

注意：距离以米为单位。

Answer 2

假设$dbStations->get('stDetails')和$dbStations->get('stationDetails')正在查询同一个表，您可以这样做：

$dbStations->select('country, _id, lat, lng, admin_level_1, locale');
$stations = $dbStations->get('stDetails');

$output = array();

foreach ($stations->result() as $station) {
    $lat2 = $station->lat;
    $lng2 = $station->lng;

    $distance = $this->distance->gpsdistance($lat1, $lng1, $lat2, $lng2, "M");

    if ($distance <= $rad) {
        $station->distance = $distance;
        $output[] = $station;
    }
}

echo json_encode($output);

我没有对此进行测试，它可能无法准确地给出您想要的内容，因此您可能需要对其进行一些调整。重要的是我是如何做到的。抓取开头的所有数据，然后验证每一行。如果行有效（在一定距离内），请将距离添加到阵列并保存以便输出。当没有剩余行时，输出有效站。

将变量附加到每个结果的末尾

2 个答案: