PHP返回NaN [英] PHP returning NaN

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);

推荐答案

您从数据库中提取的值可能是字符串，这将导致此问题.

The values you are pulling from the database may be strings, which would cause this issue.

您可能还需要检查Kolink在他的帖子中提出的问题.

You may also want to check the issues that Kolink raised in his post.

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