将raphaeljs导出为jpg且路径背景为图像时出错

Question

I have a problem with raphaeljs export plugin ( https://github.com/ElbertF/Raphael.Export ). in a path element I use attribute fill and as a source I give a image url to fullfill. But when I export this to SVG I see a path element definition, but when I export it to PNG, I do not see again.

So in my app I add an attr to path element like this:

paper.path("M 195 10 L 300  L 195 z").attr({'stroke-width': 0,'fill': 'url(images/alfen/02/murek.png)'});

and I export this with paper.toSVG()

and in my SVG I find a path:

<path transform="matrix(1,0,0,1,0,0)" fill="url(images/alfen/02/murek.png)" stroke="#000" d="M203,183.94389438943895L948,183.94389438943895L948,195L203,195Z" stroke-width="0"></path>

But when I transform this to PNG with:

<?php 
    $json        = $_POST['json'];
    $output      = str_replace('\"','"',$json);
    $filenameSVG = 'test';

    file_put_contents("$filenameSVG.svg", $output);

    $konwert = "convert $filenameSVG.svg $filenameSVG.jpg";

    system($konwert);

I cannot find this path fulfilled with my background. Can anybody help?

Answer 1

If you're able to get correct output to svg but it fails going to png in php there are several things you'll need to check.

• As a sanity check make sure your $_POST['json'] is not returning malformed json
• The next line in your php confuses me: $output = str_replace('\"','"',$json);
• It could be that the json you're returning is a single object, but I'm still not sure why you are searching the entire file and not looking for a specific nested object like $output = str_replace('\"','"',$json['filename_and_path']); and if the json you return IS a single line, there may be better ways to handle it -- i.e. post it as a string or even return each one with an array and index.

And for this stuff:

$konwert = "convert $filenameSVG.svg $filenameSVG.jpg";
system($konwert);

You may not be feeding system() with valid variables in your string. To be sure I recommend properly concatenating the string like:

$konwert = "convert".$filenameSVG.".svg ".$filenameSVG.".jpg";

You're also going to need the absolute filepath to the file on your server to execute the command on or else it won't find the file. The code $konwert = "convert".$filenameSVG.".svg ".$filenameSVG.".jpg"; is obviously only going to work for you if those two files are located in the root directory of your project.

I also don't think you should be using system() in this instance. My understanding is you should be using passthru() for dealing with image binaries. There's also exec() but really, I think what you need here is passthru(). See: http://www.php.net/manual/en/function.passthru.php