这段代码几乎可以工作
这会将整个页面输出为 jpg
问题:我怎样才能只抓取'#myDiv'
中的内容并将其输出为jpg文件?
JS:
$('.myButton').click(function(){
$('#myDiv').html2canvas();//<< has no effect
var queue = html2canvas.Parse();
var canvas = html2canvas.Renderer(queue,{elements:{length:0}});
var img = canvas.toDataURL();
img.replace(/^data:image\/(png|jpg);base64,/, "");
$.post( "postIO.php", {img:img}, function(data) {
//$('#recieve').append(data);
});
return false;
});
postIO.php:
$canvasImg = $_POST['img'];
//$canvasImg = str_replace('data:image/png;base64,', '', $canvasImg);
$data = base64_decode($canvasImg);
$File = "z.jpg";
$Handle = fopen($File, 'w');
fwrite($Handle, $data);
fclose($Handle);
引用here
最佳答案
我下载并尝试了html2canvas,然后我发现jquery插件没有完成(它只是捕获图像并创建 Canvas ,没有用),所以我自己编写捕获代码。
var el = $('div').get(0);
function saveData(dturl){
//Upload here
console.debug(dturl);
}
html2canvas.Preload(el, {
complete: function(image){
var queue = html2canvas.Parse(el, image, {elements: el}),
$canvas = $(html2canvas.Renderer(queue, {elements: el}));
saveData($canvas[0].toDataURL());
}
});
希望对你有帮助
因此要与您的程序一起使用,您必须编写
function saveData(dturl){
dturl.replace(/^data:image\/(png|jpg);base64,/, "");
$.post( "postIO.php", {img:dturl}, function(data) {
//$('#recieve').append(data);
});
}
$('.myButton').click(function(){
var el = $('#myDiv').get(0);
html2canvas.Preload(el, {
complete: function(image){
var queue = html2canvas.Parse(el, image, {elements: el}),
$canvas = $(html2canvas.Renderer(queue, {elements: el}));
saveData($canvas[0].toDataURL());
}
});
});
关于php - html2canvas 输出选定的div PHP,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10778105/