php - 如果图片存在'GPS'（纬度，经度）属性，我如何获得它们？

Question

我想从我桌面上的图片中获取经纬度信息。我在谷歌上搜索非常了解如何获取这些信息。我仍在尝试...如果有人对此有想法...那么请指导我..或如果可能的话提供任何解决方案..

Answer 1

首先确定使用 EXif：

<?php
echo "test1.jpg:<br />\n";
$exif = exif_read_data('tests/test1.jpg', 'IFD0');
echo $exif===false ? "No header data found.<br />\n" : "Image contains headers<br />\n";

$exif = exif_read_data('tests/test2.jpg', 0, true);
echo "test2.jpg:<br />\n";
foreach ($exif as $key => $section) {
    foreach ($section as $name => $val) {
        echo "$key.$name: $val<br />\n";
    }
}
?>

然后

    <?php
$image_file = 'D:\Photoes\2011\IMG_0712.jpg';

if(file_exists($image_file)){
$details = exif_read_data($image_file);
$sections = explode(',',$details['SectionsFound']);

if(in_array('GPS',array_flip($sections))){
echo format_gps_data($details['GPSLatitude'],$details['GPSLatitudeRef']);
echo '<br/>';
echo format_gps_data($details['GPSLongitude'],$details['GPSLongitudeRef']);
}else{
die('GPS data not found');
}
}else{
die('File does not exists');
}

function format_gps_data($gpsdata,$lat_lon_ref){
$gps_info = array();
foreach($gpsdata as $gps){
list($j , $k) = explode('/', $gps);
array_push($gps_info,$j/$k);
}
$coordination = $gps_info[0] + ($gps_info[1]/60.00) + ($gps_info[2]/3600.00);
return (($lat_lon_ref == "S" || $lat_lon_ref == "W" ) ? '-'.$coordination : $coordination).' '.$lat_lon_ref;
}


?>

然后看看这个： 使用 GPS INFO 的好工作