我知道已经有很多关于这方面的StackOverflow问题,但我已经搜索了我能找到的尽可能多的问题,还没有让我的代码正常工作,所以我终于发布了我自己的问题。
我的目标是将网页中HTML5 <canvas>上的图像保存到服务器上的文件中。我希望使用Java servlet来实现这一点。
我的JavasScript获取画布图像数据,如下所示:
var canvas = document.getElementById("screenshotCanvas");
var context = canvas.getContext("2d");
var imageDataURL = canvas.toDataURL('image/png');
// I'm not if I need to do this, I've tried several different ways to no avail
//imageDataURL = imageDataURL.replace("image/png", "image/octet-stream");
//imageDataURL = imageDataURL.replace(/^data:image\/(png|jpeg);base64,/,"");
$.ajax({
url: screenshotCreateUrl,
type: "POST",
data: { imgBase64: imageDataURL },
error: function(jqXHR, textStatus, errorThrown) {
// Handle errors
},
success: function(data, textStatus, jqXHR) {
// Do some stuff
}
});我的Java servlet尝试保存图像,如下所示:
try {
HttpServletRequestWrapper wrappedRequest = new HttpServletRequestWrapper(request);
HttpServletRequestWrapper requestWithWrapper = (HttpServletRequestWrapper) wrappedRequest.getRequest();
byte[] contentData = requestWithWrapper.getContentData();
byte[] decodedData = Base64.decodeBase64(contentData);
FileOutputStream fos = new FileOutputStream("testOutput.png");
fos.write(decodedData);
fos.close();
} catch(Exception e) {
// Handle exceptions
}servlet成功地写出了一个图像文件,但它没有正确打开,并且没有包含其中的所有图像数据。我的Javascript成功获取了<canvas>图像数据,如下所示:
data:image/png;base64,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 . . . [and so on]你知道我这里遗漏了什么吗?我觉得我犯了一些小错误,我就是看不出来。
发布于 2013-12-31 02:00:32
有相同的任务,并能够通过使用multipart/form-data内容类型来使其工作(没有jQuery,并在maclema的帮助下):
var xhr = new XMLHttpRequest();
xhr.open("post", "AddressOfYourServlet", false);
var boundary = Math.random().toString().substr(2);
xhr.setRequestHeader("content-type",
"multipart/form-data; charset=utf-8; boundary=" + boundary);
var multipart = "--" + boundary + "\r\n" +
"Content-Disposition: form-data; name=myImg\r\n" +
"Content-type: image/png\r\n\r\n" +
canvas.toDataURL("image/png") + "\r\n" +
"--" + boundary + "--\r\n";
xhr.send(multipart);要异步执行,或者如果您有更多部分要发送(例如,多个图像),或者如果您希望处理响应,请参阅How to send multipart/form-data form content by ajax (no jquery)?
servlet的doPost方法将如下所示:
protected void doPost(HttpServletRequest request, HttpServletResponse response)
throws ServletException, IOException {
Part part = request.getPart("myImg");
BufferedReader br = new BufferedReader(new InputStreamReader(part.getInputStream(),
Charset.forName("utf-8")));
String sImg = br.readLine();
sImg = sImg.substring("data:image/png;base64,".length());
byte[] bImg64 = sImg.getBytes();
byte[] bImg = Base64.decodeBase64(bImg64); // apache-commons-codec
FileOutputStream fos = new FileOutputStream("img.png");
fos.write(bImg);
}希望这能有所帮助。
发布于 2013-06-19 06:29:17
您希望获取post参数,而不是请求的内容数据。此外,您还需要剥离编码信息。
试试这个:
try {
HttpServletRequestWrapper wrappedRequest = new HttpServletRequestWrapper(request);
HttpServletRequestWrapper requestWithWrapper = (HttpServletRequestWrapper) wrappedRequest.getRequest();
String imageString = wrappedRequest.getParameter("imgBase64");
imageString = imageString.substring("data:image/png;base64,".length);
byte[] contentData = imageString.getBytes();
byte[] decodedData = Base64.decodeBase64( contentData );
FileOutputStream fos = new FileOutputStream("testOutput.png");
fos.write(decodedData);
fos.close();
} catch(Exception e) {
// Handle exceptions
e.printStackTrace();
}https://stackoverflow.com/questions/17178779
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