PHP提取GPS EXIF数据 [英] PHP extract GPS EXIF data

问题描述

我想从使用php的图片中提取GPS EXIF标签。
我使用 exif_read_data（）返回所有标签+数据的数组：

<$ p $ [p] GPS.GPSLatitudeRef：N
GPS.GPSLatitude：Array（[0] => 46/1 [1] => 5403/100 [2] => 0 / 1）
GPS.GPSLongitudeRef：E
GPS.GPSLongitude：Array（[0] => 7/1 [1] => 880/100 [2] => 0/1）
GPS.GPSAltitudeRef：
GPS.GPSAltitude：634/1


I don'不知道如何解释46/1 5403/100和0/1？ 46可能是46°，但其余的尤其是0/1呢？

  angle / 1 5403/100 0/1 
  

这是什么结构？

如何将它们转换为标准（例如来自维基百科的46°56'48N 7°26'39E）？我想通过坐标到谷歌地图API来显示地图上的图片位置！ 解决方案

根据 http://en.wikipedia.org/wiki/Geotagging ，（[0 ] => 46/1 [1] => 5403/100 [2] => 0/1）应该表示46/1度，5403/100分钟，0/1秒，即46°54.03'0N。正常化的秒数为46°54'1.8N。

只要你没有得到负坐标（假设你得到N / S和E / W作为单独的坐标，您不应该有负坐标）。

  //通过在GPS.GPSLatitude或GPS.GPSLongitude或以这种格式的东西
 function getGps（$ exifCoord）
 {
 $ degrees = count（$ exifCoord）> 0？ gps2Num（$ exifCoord [0]）：0; 
 $分钟= count（$ exifCoord）> 1？ gps2Num（$ exifCoord [1]）：0; 
 $ seconds = count（$ exifCoord）> 2？ gps2Num（$ exifCoord [2]）：0; 
 
 //标准化
 $分钟+ = 60 *（$ degrees  -  floor（$ degrees））; 
 $ degrees = floor（$ degrees）; 
 
 $秒+ = 60 *（$分钟 - 楼层（$分钟））; 
 $分钟=楼层（$分钟）; 
 
 //额外的标准化，可能不需要，除非你得到奇怪的数据
 if（$ seconds> = 60）
 {
 $ minutes + = floor（$秒/ 60.0）; 
 $ seconds  -  = 60 * floor（$ seconds / 60.0）; （$分钟> = 60）
 {
 $ degrees + = floor（$分钟/ 60.0）; 
} 
 
 
 $分钟 -  = 60 *地板（$分钟/ 60.0）; 
} 
 
 return array（'degrees'=> $ degrees，'minutes'=> $ minutes，'seconds'=> $ seconds）; 
} 
 
函数gps2Num（$ coordPart）
 {
 $ parts = explode（'/'，$ coordPart）; 
 
 if（count（$ parts）< = 0）// jic 
 return 0; 
 if（count（$ parts）== 1）
 return $ parts [0]; 
 $ b $返回floatval（$ parts [0]）/ floatval（$ parts [1]）; 
} 
  

I would like to extract the GPS EXIF tag from pictures using php. I'm using the exif_read_data() that returns a array of all tags + data :

GPS.GPSLatitudeRef: N
GPS.GPSLatitude:Array ( [0] => 46/1 [1] => 5403/100 [2] => 0/1 ) 
GPS.GPSLongitudeRef: E
GPS.GPSLongitude:Array ( [0] => 7/1 [1] => 880/100 [2] => 0/1 ) 
GPS.GPSAltitudeRef: 
GPS.GPSAltitude: 634/1

I don't know how to interpret 46/1 5403/100 and 0/1 ? 46 might be 46° but what about the rest especially 0/1 ?

angle/1 5403/100 0/1

What is this structure about ?

How to convert them to "standard" ones (like 46°56′48″N 7°26′39″E from wikipedia) ? I would like to pass thoses coordinates to the google maps api to display the pictures positions on a map !

解决方案

According to http://en.wikipedia.org/wiki/Geotagging, ( [0] => 46/1 [1] => 5403/100 [2] => 0/1 ) should mean 46/1 degrees, 5403/100 minutes, 0/1 seconds, i.e. 46°54.03′0″N. Normalizing the seconds gives 46°54′1.8″N.

This code below should work, as long as you don't get negative coordinates (given that you get N/S and E/W as a separate coordinate, you shouldn't ever have negative coordinates). Let me know if there is a bug (I don't have a PHP environment handy at the moment).

//Pass in GPS.GPSLatitude or GPS.GPSLongitude or something in that format
function getGps($exifCoord)
{
  $degrees = count($exifCoord) > 0 ? gps2Num($exifCoord[0]) : 0;
  $minutes = count($exifCoord) > 1 ? gps2Num($exifCoord[1]) : 0;
  $seconds = count($exifCoord) > 2 ? gps2Num($exifCoord[2]) : 0;

  //normalize
  $minutes += 60 * ($degrees - floor($degrees));
  $degrees = floor($degrees);

  $seconds += 60 * ($minutes - floor($minutes));
  $minutes = floor($minutes);

  //extra normalization, probably not necessary unless you get weird data
  if($seconds >= 60)
  {
    $minutes += floor($seconds/60.0);
    $seconds -= 60*floor($seconds/60.0);
  }

  if($minutes >= 60)
  {
    $degrees += floor($minutes/60.0);
    $minutes -= 60*floor($minutes/60.0);
  }

  return array('degrees' => $degrees, 'minutes' => $minutes, 'seconds' => $seconds);
}

function gps2Num($coordPart)
{
  $parts = explode('/', $coordPart);

  if(count($parts) <= 0)// jic
    return 0;
  if(count($parts) == 1)
    return $parts[0];

  return floatval($parts[0]) / floatval($parts[1]);
}

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