Question

我一直试图从照片中显示一些基本的EXIF数据而没有运气。我的代码如下。我只需要显示一些基本的信息。我在服务器上启用了EXIF，我知道此图像中有EXIF数据 - 但是没有数据显示此设置。谁能告诉我这里我做错了什么？

$exif_data = exif_read_data("http://www.findexif.com/client/samples/swiss_house.jpg");

        $emake = $exif_data['Make'];
        $emodel = $exif_data['Model'];
        $eexposuretime = $exif_data['ExposureTime'];
        $efnumber = $exif_data['FNumber'];
        $eiso = $exif_data['ISOSpeedRatings'];
        $edate = $exif_data['DateTime'];

    echo "Make: ". $emake . "<br>";
        echo "Model: ". $emodel . "<br>";
        echo "Exp: ". $eexposuretime . "<br>";
        echo "Date: ". $edate . "<br>";
        echo "ISO: ". $eiso . "<br>";

Answer 1

exif_read_data的文件名不能是网址。

如果您仍想阅读外部图片的EXIF数据，则需要先下载：

<?php
$imageURL = 'http://www.findexif.com/client/samples/swiss_house.jpg';
$localURL = '/tmp/swiss_house.jpeg';

// Use 'copy' function if you have PHP5
//  and the HTTP stream wrapper enabled on your server.
copy($imageURL, $localURL);

/**
 * This way if you don't have PHP5
$fp = fopen($localURL, 'w');
fwrite($fp, file_get_contents($imageURL));
fclose($fp);
 */

$exif_data = exif_read_data($localURL);

$emake = $exif_data['Make'];
$emodel = $exif_data['Model'];
$eexposuretime = $exif_data['ExposureTime'];
$efnumber = $exif_data['FNumber'];
$eiso = $exif_data['ISOSpeedRatings'];
$edate = $exif_data['DateTime'];

echo 'Make: '. $emake . '<br>';
echo 'Model: '. $emodel . '<br>';
echo 'Exp: '. $eexposuretime . '<br>';
echo 'Date: '. $edate . '<br>';
echo 'ISO: '. $eiso . '<br>';

PHP EXIF数据无法显示

1 个答案: