Convert string to image with PHP

Question

I have a PHP script that should create a valid image file on server side. The following code runs fine:

$filename = $_GET['filename'];

// Only proceed if we got valid input
if ($filename !== null) {
    echo "$filename is not null.";
    $image = @imagecreatetruecolor(10, 10)
                or die('Cannot Initialize new GD image stream');

    if (strpos($image, '/gif') !== false) {
        $image_type = "GIF";
        header('Content-Type: image/gif');
        $successful = imagegif($image, "./$filename");
    } else if (strpos($image, '/jpeg') !== false) {
        $image_type = "JPG";
        header('Content-Type: image/jpeg');
        $successful = imagejpeg($image, "./$filename");
    } else if (strpos($image, '/png') !== false) {
        $image_type = "PNG";
        header('Content-Type: image/png');
        $successful = imagepng($image, "./$filename");
    }

    if ($successful) {
        echo "Image written to '$filename'.";
    } else {
        echo "Could not write $image_type image to '$filename'.";
    }
    imagedestroy($image);
    echo "image destroyed.";
} else {
    echo "$filename is null.";
}

This works fine & an image with the $filename is created. But actually, I have to receive not only the file name but the image also. So, the real code is

$filename = $_GET['filename'];

$image = $_GET['image'];

echo "file $filename = '$image'.";

// Only proceed if we got valid input
if ($filename !== null) {
    echo "$filename is not null.";

    if (strpos($image, '/gif') !== false) {
        $image_type = "GIF";
        header('Content-Type: image/gif');
        $successful = imagegif($image, "./$filename");
    } else if (strpos($image, '/jpeg') !== false) {
        $image_type = "JPG";
        header('Content-Type: image/jpeg');
        $successful = imagejpeg($image, "./$filename");
    } else if (strpos($image, '/png') !== false) {
        $image_type = "PNG";
        header('Content-Type: image/png');
        $successful = imagepng($image, "./$filename");
    }

    if ($successful) {
        echo "Image written to '$filename'.";
    } else {
        echo "Could not write $image_type image to '$filename'.";
    }
    imagedestroy($image);
    echo "image destroyed.";
} else {
    echo "$filename is null.";
}

This does not work, and the result is

file t.png = 'data:image/jpeg;base64,/9j/4AAQSkZJ … qA/Cz//Z'.t.png is not null.Image written to 't.png'.image destroyed.

How can I create from the 'data:image/jpeg … ' string a valid image in PHP?

EDIT 1: I added one line to the code above to see that it is not a possible duplicate of another question:

// Only proceed if we got valid input
if ($filename !== null) {
    echo "$filename is not null.";
    $image = base64_decode($image); // <<<<

EDIT 2: I modified the code so that the file is deleted:

$filename = $_GET['filename'];

$image = $_GET['image'];

// Only proceed if we got valid input
if ($filename !== null) {
    echo "$filename is not null.";
    $image = base64_decode($image);
    $slash1 = strpos($image, '/');

    $image_type = substr($image, $slash1, strpos($image, ';') - $slash1);

    if (file_exists($filename)) unlink($filename);

    header('Content-Type: image/' . $image_type);

    switch ($image_type) {
        case "gif":
            $successful = imagegif($image, "./$filename");
            break;
        case "jpeg":
        case "jpg":
            $successful = imagejpeg($image, "./$filename");
            break;
        case "png":
            $successful = imagepng($image, "./$filename");
            break;
    }

    if ($successful) {
        echo "Image written to '$filename'.";
    } else {
        echo "Could not write $image_type image to '$filename'.";
    }
    imagedestroy($image);
    echo "image destroyed.";
} else {
    echo "$filename is null.";
}

But still, I get the reply Could not write image to 't.jpg'.

EDIT 3: This is what I pass to the two parameters:

?filename=t.jpg&image=data:image/jpeg;base64,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

EDIT 4: Modified the code according to suggestions of @delboy1978uk. Also, it is not correct that the passed filename includes the image file extension, as this should always correspond to the image type:

$filename = $_GET['filename'];

$image = $_GET['image'];

echo "file $filename = '$image'.";

// Only proceed if we got valid input
if ($filename !== null) {
    $slash = strpos($image, '/') + 1;

    $image_type = substr($image, $slash, strpos($image, ';') - $slash);
    $comma = strpos($image, ',') + 1;

    $image = substr($image, $comma);

    $decoded_image = base64_decode($image);

    $image = imagecreatefromstring($decoded_image);

    echo "The image type is '$image_type'.";

    if (file_exists($filename)) {
        unlink($filename);
        echo "Deleted file '$filename'.";
    }

    header('Content-Type: image/' . $image_type);
    $filename .= '.' . $image_type;

    switch ($image_type) {
        case "gif":
            $successful = imagegif($image, "./$filename");
            break;
        case "jpeg":
        case "jpg":
            $successful = imagejpeg($image, "./$filename");
            break;
        case "png":
            $successful = imagepng($image, "./$filename");
            break;
    }

    if ($successful) {
        echo "Image written to '$filename'.";
    } else {
        echo "Could not write $image_type image to '$filename'.";
    }

    if (imagedestroy($image) === true) {
        echo "Image destroyed.";
    }
} else {
    echo "$filename is null.";
}

Possible duplicate of Create Image from Data URL

iainn
– iainn

2017-09-19 11:28:32 +00:00
Commented Sep 19, 2017 at 11:28 — iainn
– iainn, Commented Sep 19, 2017 at 11:28

delboy1978uk · Accepted Answer · 2017-09-19 11:52:05Z

2

You are trying to send headers on a page you have already sent output to.

If you are loading from a file, You need to use imagecreatefromjpeg() and the GIF and PNG equivalents. http://php.net/manual/en/function.imagecreatefromjpeg.php

$img = imagecreatefromjpeg($file);

To get the actual string data to literally echo in your <image> tag, use output buffering:

ob_start();
imagejpeg($img)
$image = ob_get_clean();
echo '<img src="data:image/jpeg;base64,' . base64_encode( $i ).'" />';

If you are interested, I made an Image class a few years ago that handles this stuff, see here https://github.com/delboy1978uk/image/blob/master/src/Image.php and a blog about it https://delboy1978uk.wordpress.com/2014/12/01/outputting-images-as-base64-encoded-strings/

answered Sep 19, 2017 at 11:52

delboy1978uk

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8 Comments

Sae1962 Over a year ago

Oh, thank you! But imagecreatefromjpeg needs a file name (of an image file) to run, and I am sending text data with a GET request from which I am supposed to create a file on the server disk. I.e. I want to duplicate the file created on the client side on the server side.

delboy1978uk Over a year ago

Ah ok, I thought in the second part they were sending a filename for you to open! In that case, it sounds like it is failing when the file already exists? try file_exists() and maybe remove it if it does?

Sae1962 Over a year ago

Modified the code, but there is something wrong, as no file is created.

delboy1978uk Over a year ago

$image = $_GET['image']; - What is this then? Your working code doesn't use it, but your broken code does. Remember everything in a GET query will be url encoded? Maybe switch to post?

Sae1962 Over a year ago

Ehmm. Good point, but I already switched to GET with a tiny, tiny image so that I can debug the whole issue. Actually, in this example, a GET is the only option, as I do not use any client side. The real code will have to use POST for sure, but then, I cannot see what really is the matter. But my code uses $image a lot of times, doesn't it?

|

Sae1962 · Accepted Answer · 2017-09-21 11:31:39Z

The problem is solved!! Here is the code that gets the browser-cached image & sends it to a requested server location. Getting back the image from the server, it turns out to be the same image. First the JavaScript code:

[…]
var parameters = {
    […],
    headerImageLocation: '',
    image: new Image(),
    […],
};

if (strings.hasMinimalLength(blobURL, 9)) {
    this.getBlobFromURL(blobURL).then(this.fromBlobToBase64).then(function(result) {
        parameters['image'].src = result;
        parameters['headerImageLocation'] = './' + strings.generateID();

        server.continueWithNewsTicker(parameters);
    });
} else {
    this.continueWithNewsTicker(parameters);
}

// Prototype "MainServer":
MainServer.method('continueWithNewsTicker', function(parameters) {
    var url = server.ServiceTest + 'saveHeaderImage.php';

    if (strings.hasMinimalLength(parameters['headerImageLocation'], 1)) {
        var formData = new FormData();

        formData.append('filename', parameters['headerImageLocation']);
        formData.append('image', parameters['image'].src);

        this.uploadFile(formData, url);
    }
    […]
});

// Prototype "Server":
Server.method('uploadFile', function (data, url) {
    var xhr = new XMLHttpRequest();    // AJAX request

    xhr.open('POST', url);
    xhr.send(data);
});

// Prototype "Strings":
Strings.method('generateID', function () {
    function s4() {
        return Math.floor((1 + Math.random())*0x10000)
            .toString(16)
            .substring(1);
    }
    return s4() + s4() + '-' + s4() + '-' + s4() + '-' +
        s4() + '-' + s4() + s4() + s4();
});

For those interested, the PHP code follows:

ini_set('display_errors', 1);
error_reporting(E_ALL);

// Get the input data safely
$filename = $_POST['filename'];
$image = $_POST['image'];

// Only proceed if we got valid input
if ($filename !== null) {
    // Prepare to remove "header" information:
    $comma = strpos($image, ',') + 1;
    $slash = strpos($image, '/') + 1;
    // The image type will also determine the file extension
    $image_type = substr($image, $slash, strpos($image, ';') - $slash);
    // Remove "header" information from rest of image:
    $image = substr($image, $comma);

    $decoded_image = base64_decode($image);
    $filename .= '.' . $image_type;

    if (file_exists($filename)) unlink($filename);

    $successful = file_put_contents($filename, $decoded_image);

    if ($successful) {
        echo "Image written to '$filename'.";
    } else {
        echo "Could not write $image_type image to '$filename'.";
    }
} else {
    echo "$filename is null.";
}

Little tip for you, displaying errors kills any header calls and ajax functionality if there is any output. For the best error logging experience, set error_reporting to -1, turn display_errors off, and set a custom error_log. Then in the terminal, type 'tail -f /path/to/error_log'. Your notices, warnings and errors will now scroll past in real time, without distorting your web page's display.

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