使用PHP从JPG中删除EXIF数据

Is there any way to remove the EXIF data from a JPG using PHP? I have heard of PEL, but I'm hoping there's a simpler way. I am uploading images that will be displayed online and would like the EXIF data removed.

有没有办法使用PHP从JPG中删除EXIF数据?我听说过PEL,但我希望有一种更简单的方法。我正在上传将在线显示的图片,并希望删除EXIF数据。

Thanks!

EDIT: I don't/can't install ImageMagick.

编辑:我不能/不能安装ImageMagick。

8 个解决方案

#1

Use gd to recreate the graphical part of the image in a new one, that you save with another name.

使用gd以新的名称重新创建图像的图形部分,并使用其他名称保存。

See PHP gd

请参阅PHP gd

edit 2017

Use the new Imagick feature.

使用新的Imagick功能。

Open Image:

<?php
    $incoming_file = '/Users/John/Desktop/file_loco.jpg';
    $img = new Imagick(realpath($incoming_file));

Be sure to keep any ICC profile in the image

请务必在图像中保留任何ICC配置文件

    $profiles = $img->getImageProfiles("icc", true);

then strip image, and put the profile back if any

然后剥离图像,然后将配置文件放回去

    $img->stripImage();

    if(!empty($profiles)) {
       $img->profileImage("icc", $profiles['icc']);
    }

Comes from this PHP page, see comment from Max Eremin down the page.

来自这个PHP页面,请参阅Max Eremin的评论。

#2

A fast way to do it in PHP using ImageMagick (Assuming you have it installed and enabled).

使用ImageMagick在PHP中快速完成此操作(假设您已安装并启用它)。

<?php

$images = glob('*.jpg');

foreach($images as $image) 
{   
    try
    {   
        $img = new Imagick($image);
        $img->stripImage();
        $img->writeImage($image);
        $img->clear();
        $img->destroy();

        echo "Removed EXIF data from $image. \n";

    } catch(Exception $e) {
        echo 'Exception caught: ',  $e->getMessage(), "\n";
    }   
}
?>

#3

I was looking for a solution to this as well. In the end I used PHP to rewrite the JPEG with ALL Exif data removed. I didn't need any of it for my purposes.

我也在寻找解决方案。最后,我使用PHP重写了JPEG,删除了所有Exif数据。我不需要任何这个用于我的目的。

This option has several advantages...

这个选项有几个优点......

The file is smaller because the EXIF data is gone.

该文件较小,因为EXIF数据消失了。

There is no loss of image quality (because the image data is unchanged).

不会损失图像质量(因为图像数据不变)。

Also a note on using imagecreatefromjpeg: I tried this and my files got bigger. If you set quality to 100, your file will be LARGER, because the image has been resampled, and then stored in a lossless way. And if you don't use quality 100, you lose image quality. The ONLY way to avoid resampling is to not use imagecreatefromjpeg.

还有关于使用imagecreatefromjpeg的说明:我试过这个,我的文件变大了。如果将质量设置为100,则文件将更大,因为图像已经过重新采样,然后以无损方式存储。如果您不使用质量100,则会丢失图像质量。避免重新采样的唯一方法是不使用imagecreatefromjpeg。

Here is my function...

这是我的功能......

function removeExif($old, $new)
{
    $f1 = fopen($old, 'rb');
    $f2 = fopen($new, 'wb');

    // Find EXIF marker
    while (($s = fread($f1, 2))) {
        $word = unpack('ni', $s)['i'];
        if ($word == 0xFFE1) {
            // Read length (includes the word used for the length)
            $s = fread($f1, 2);
            $len = unpack('ni', $s)['i'];
            // Skip the EXIF info
            fread($f1, $len - 2);
            break;
        } else {
            fwrite($f2, $s, 2);
        }
    }

    // Write the rest of the file
    while (($s = fread($f1, 4096))) {
        fwrite($f2, $s, strlen($s));
    }

    fclose($f1);
    fclose($f2);
}

#4

The following will remove all EXIF data of a jpeg file. This will make a copy of original file without EXIF and remove the old file. Use 100 quality not to loose any quality details of picture.

以下将删除jpeg文件的所有EXIF数据。这将生成没有EXIF的原始文件的副本并删除旧文件。使用100质量不会丢失图片的任何质量细节。

$path = "/image.jpg";

$img = imagecreatefromjpeg ($path);
imagejpeg ($img, $path, 100);
imagedestroy ($img);

(simple approximation to the graph can be found here)

(图表的简单近似可以在这里找到)

#5

I completely misunderstood your question.

我完全误解了你的问题。

You could use some command line tool to do this job. or write your own php extension to do it. have a look at this lib that would be useful: http://www.sno.phy.queensu.ca/~phil/exiftool/

您可以使用一些命令行工具来完成这项工作。或者编写自己的php扩展来完成它。看看这个有用的库:http://www.sno.phy.queensu.ca/~phil/exiftool/

Cheers,

vfn

#6

I'm not pretty sure about it, but if its possible using GD o ImageMagick, the first thing that come to my mind is to create a new Image and add the old image to the new one.

我不太确定它,但如果它可能使用GD o ImageMagick,我想到的第一件事是创建一个新的图像并将旧图像添加到新图像。

#7

this is the simplest way:

这是最简单的方法:

$images = glob($location.'/*.jpg');

foreach($images as $image) {   
    $img = imagecreatefromjpeg($image);
    imagejpeg($img,$image,100);
}

#8

function remove_exif($in, $out)
{
    $buffer_len = 4096;
    $fd_in = fopen($in, 'rb');
    $fd_out = fopen($out, 'wb');
    while (($buffer = fread($fd_in, $buffer_len)))
    {
        //  \xFF\xE1\xHH\xLLExif\x00\x00 - Exif 
        //  \xFF\xE1\xHH\xLLhttp://      - XMP
        //  \xFF\xE2\xHH\xLLICC_PROFILE  - ICC
        //  \xFF\xED\xHH\xLLPhotoshop    - PH
        while (preg_match('/\xFF[\xE1\xE2\xED\xEE](.)(.)(exif|photoshop|http:|icc_profile|adobe)/si', $buffer, $match, PREG_OFFSET_CAPTURE))
        {
            echo "found: '{$match[3][0]}' marker\n";
            $len = ord($match[1][0]) * 256 + ord($match[2][0]);
            echo "length: {$len} bytes\n";
            echo "write: {$match[0][1]} bytes to output file\n";
            fwrite($fd_out, substr($buffer, 0, $match[0][1]));
            $filepos = $match[0][1] + 2 + $len - strlen($buffer);
            fseek($fd_in, $filepos, SEEK_CUR);
            echo "seek to: ".ftell($fd_in)."\n";
            $buffer = fread($fd_in, $buffer_len);
        }
        echo "write: ".strlen($buffer)." bytes to output file\n";
        fwrite($fd_out, $buffer, strlen($buffer));
    }
    fclose($fd_out);
    fclose($fd_in);
}

It is a prototype for a call from a command line.

它是来自命令行的调用的原型。

#1