PHP返回NaN

本文介绍了PHP返回NaN的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我有一个计算两个GPS坐标之间距离的函数.然后，我从数据库中获取所有坐标并遍历它们，以获取当前坐标与上一个坐标之间的距离，然后将其添加到特定GPS设备的数组中.由于某种原因，它返回NaN.我尝试将其强制转换为双精度，整数并四舍五入.

I have a function that calculates the distance between two GPS coordinates. I then get all the coordinates from the database and loop through them all to get the distance between the current one and the previous one, then add that to an array for the specific GPS device. For some reason it is return NaN. I have tried casting it as a double, an int, and rounding the number.

这是我的PHP代码:

function distance($lat1, $lon1, $lat2, $lon2) {
      $lat1 = round($lat1, 3);
      $lon1 = round($lon1, 3);
      $lat2 = round($lat2, 3);
      $lon2 = round($lon2, 3);
      $theta = $lon1 - $lon2;
      $dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta));
      $dist = acos($dist);
      $dist = rad2deg($dist);
      $miles = $dist * 60 * 1.1515;
      if($miles < 0) $miles = $miles * -1;
      return ($miles * 1.609344);
}
$this->db->query("SELECT * FROM `gps_loc` WHERE `imeiN`='" . $sql . "' AND `updatetime`>=$timeLimit ORDER BY `_id` DESC");
    $dist = array();
    $dist2 = array();
    while($row = $this->db->getResults()) {
        $dist2[$row['imeiN']] = 0;
        $dist[$row['imeiN']][]["lat"] = $row['lat'];
        $dist[$row['imeiN']][count($dist[$row['imeiN']]) - 1]["lng"] = $row['lon'];
    }

    foreach($dist as $key=>$d) {
        $a = 0;
        $b = 0;
        foreach($dist[$key] as $n) {
            if($a > 0) {
                $dist2[$key] += $this->distance($n['lat'], $n['lng'], $dist[$key][$a - 1]['lat'], $dist[$key][$a - 1]['lng']);
            }
            $a++;
        }

    }
    echo json_encode($dist2);

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