Question

我在准备好的声明中返回$ distance时遇到了一些问题。 $ station_id被呼出，一切都很好，但距离是空的。

我还替换了问号并直接在phpmyadmin中运行查询，它返回distance和station_id列，以便查询本身显示正常。

难以弄清楚原因，并想知道是否有人可以解决出错的问题。请找到下面的代码，如果您有任何疑问，请不要犹豫。哦，对于背景，sql公式是haversine formula

非常感谢任何帮助，再次感谢。

 <?php
   require_once("../includes/station.php");
   require_once("../includes/maps.php");

   global $conn;

   $latitude = 54.335;
   $longitude = -1.45007;
   //$distance = 50;

   $limit_start = 0;
   $limit_end = 5;

   //$query = "SELECT *,(((acos(sin((".$latitude."*pi()/180)) * sin((`Latitude`*pi()/180))+cos((".$latitude."*pi()/180)) * cos((`Latitude`*pi()/180)) * cos(((".$longitude."- `Longitude`)*pi()/180))))*180/pi())*60*1.1515) as distance FROM `postcodelatlng` having distance <= ".$distance;
   //$query .= ' order by distance asc limit 20, 40';

   //works
   //SELECT station_details.station_id, (((acos(sin((54.335*pi()/180)) * sin((`Latitude`*pi()/180))+cos((54.335*pi()/180)) * cos((`Latitude`*pi()/180)) * cos(((-1.45007- `Longitude`)*pi()/180))))*180/pi())*60*1.1515) as distance FROM `postcodelatlng`inner join station_details on station_details.post_code = postcodelatlng.postcode

   // Query uses the haversine formula to calculate distance in miles as the crow flies
   $query = "SELECT station_details.station_id,
          (((acos(sin((?*pi()/180)) * sin((`Latitude`*pi()/180))+cos((?*pi()/180)) * cos((`Latitude`*pi()/180)) * cos(((? - `Longitude`)*pi()/180))))*180/pi())*60*1.1515) AS distance
   FROM `postcodelatlng`
   INNER JOIN station_details ON station_details.post_code = postcodelatlng.postcode
   INNER JOIN core ON station_details.station_id = core.entity_id
   WHERE core.live = 1
   ORDER BY distance ASC";

   $query .= ' limit ?, ?';

   //echo $query;

   if ($stmt = $conn->prepare($query)) {
           $stmt->bind_param('sssii', $lattitude, $lattitude, $longitude, $limit_start, $limit_end);
           $stmt->execute();

           $stmt->bind_result($station_id, $distance);

           while ($stmt->fetch()) {
              echo "id: $station_id dist: $distance";
           }

           $stmt->close();
   }
   $conn->close();
   ?>

mysqli准备好的陈述 - 并非所有结果都具有约束力

0 个答案: