Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
736 views
in Technique[技术] by (71.8m points)

metadata - PHP extract GPS EXIF data

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 !

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

This is my modified version. The other ones didn't work for me. It will give you the decimal versions of the GPS coordinates.

The code to process the EXIF data:

$exif = exif_read_data($filename);
$lon = getGps($exif["GPSLongitude"], $exif['GPSLongitudeRef']);
$lat = getGps($exif["GPSLatitude"], $exif['GPSLatitudeRef']);
var_dump($lat, $lon);

Prints out in this format:

float(-33.8751666667)
float(151.207166667)

Here are the functions:

function getGps($exifCoord, $hemi) {

    $degrees = count($exifCoord) > 0 ? gps2Num($exifCoord[0]) : 0;
    $minutes = count($exifCoord) > 1 ? gps2Num($exifCoord[1]) : 0;
    $seconds = count($exifCoord) > 2 ? gps2Num($exifCoord[2]) : 0;

    $flip = ($hemi == 'W' or $hemi == 'S') ? -1 : 1;

    return $flip * ($degrees + $minutes / 60 + $seconds / 3600);

}

function gps2Num($coordPart) {

    $parts = explode('/', $coordPart);

    if (count($parts) <= 0)
        return 0;

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

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

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...