CODEWITHSUNDEEP
phpajaximagecanvastodataurl
compilingJohnny
compilingJohnny

Reputation: 107

Saving a base 64 image creates a blank image in PHP

First here's my client side code: 

$("#fileToUpload").on("change", function(){
            var filesToUpload = document.getElementById("fileToUpload");
            var file = filesToUpload.files[0];
            var img = new Image(600,400);
            var reader = new FileReader();
            reader.onload = function(e){
                img.src = e.target.result;
            }
            reader.readAsDataURL(file);

            var canvas = document.getElementById("canvas");
            var ctx = canvas.getContext('2d');
            img.onload = function(){

                canvas.width = 600;
                canvas.height = 400;

                ctx.drawImage(img,0,0,canvas.width,canvas.height);
            }

            var dataURL = canvas.toDataURL("image/jpg");

                var data = new FormData();
                data.append("image", dataURL);

                var xhttp = new XMLHttpRequest;
                xhttp.open("POST", "test.php", true);

                xhttp.send(data);

        })

And here's my php code: 

$imageArr = explode(',', $_POST['image']);
$image = base64_decode($imageArr[1]);
file_put_contents('image.jpg',$image);

I'm resizing the image on the client side and then sending it as a data url to php to then be saved as an image.

When I save an image it creates a blank image on my server. BUT when I save another image without refreshing the page it saves the previous image in it's place and this time correctly. This is beyond me, can someone please shed some light?

Upvotes: 2

Views: 266

Answers (2)

Emil Alkalay
Emil Alkalay

Reputation: 496

You are sending the image before it is loaded. Your img.onload function is executed after xhttp.send(data). When you upload one more time it gets previous image which is already loaded.

Try following:

$("#fileToUpload").on("change", function(){
        var filesToUpload = document.getElementById("fileToUpload");
        var file = filesToUpload.files[0];
        var img = new Image(600,400);
        var reader = new FileReader();
        reader.onload = function(e){
            img.src = e.target.result;
        }
        reader.readAsDataURL(file);

        var canvas = document.getElementById("canvas");
        canvas.width = 600;
        canvas.height = 400;
        var ctx = canvas.getContext('2d');

        img.onload = function(){

            ctx.drawImage(img,0,0,canvas.width,canvas.height);

            var dataURL = canvas.toDataURL("image/jpg");

            var data = new FormData();
            data.append("image", dataURL);

            var xhttp = new XMLHttpRequest;
            xhttp.open("POST", "test.php", true);

            xhttp.send(data);
        }

    })

Upvotes: 1

Martin Öjes
Martin Öjes

Reputation: 91

A possibility to keep in mind: Since everything on client side is setup in the file input's onChange event handler, maybe some setup code is launched in the wrong order and/or not in time to be used as intended? Then, when invoking the onChange event again, resources are already in place /has already been initialized and the image gets displayed as intended.

It's just a theory. To investigate, try to move initializations out of the event handler scope.

Upvotes: 0

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