我将一些图像加载到我的画布上，然后在加载后我想单击一个按钮将该画布图像保存到我的服务器上。我可以看到脚本工作正常，直到它到达“toDataURL”部分并且我的函数停止执行。我究竟做错了什么？这是我的代码：
<!DOCTYPE HTML>
<html>
<head>
<style>
body {
margin: 0px;
padding: 0px;
}
</style>
</head>
<body>
<canvas id="myCanvas" width="578"
height="200"></canvas>
<div>
<button onClick="saveCards();">Save</button>
</div>
<script>
function loadImages(sources, callback)
{
var images = {};
var loadedImages = 0;
var numImages = 0;
// get num of sources
for(var src in sources) {
numImages++;
}
for(var src in sources) {
images[src] = new Image();
images[src].onload = function() {
if(++loadedImages >= numImages)
{
callback(images);
}
};
images[src].src = sources[src];
}
}
var canvas =
document.getElementById('myCanvas');
var context = canvas.getContext('2d');
var sources = {
great:
'images/great.jpg',
star:
'images/1Star.jpg', good:
'images/good.jpg'
};
loadImages(sources, function(images) {
context.drawImage(images.great,
0, 0, 80, 120);
context.drawImage(images.star, 80,
0, 80, 120);
context.drawImage(images.good, 160, 0, 80,
120);
});
</script>
<script type="text/javascript">
function saveCards()
{
var canvas=
document.getElementById("myCanvas");
alert("stops");
var theString= canvas.toDataURL();
var postData= "CanvasData="+theString;
var ajax= new XMLHttpRequest();
ajax.open("POST", 'saveCards.php', true);
ajax.setRequestHeader('Content-Type',
'canvas/upload');
ajax.onreadystatechange=function()
{
if(ajax.readyState == 4)
{
alert("image was saved");
}else{
alert("image was not saved");
}
}
ajax.send(postData);
}
</script>
</body>
</html>
感谢您的帮助，是因为在调用 toDataUrl 之前未加载图像吗？如果是这样，你能帮我解决它吗？
这是 PHP 脚本：
<?php
if(isset($GLOBALS['HTTP_RAW_POST_DATA']));
{
$rawImage=$GLOBALS['HTTP_RAW_POST_DATA'];
$removeHeaders=
substr($rawImage,strpos($rawImage, ",")+1);
$decode=base64_decode($removeHeaders);
$fopen= fopen('images/image.png', 'wb');
fwrite($fopen, $decode);
fclose($fopen);
}
?>
不过我遇到了安全错误。
