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使用Java Servlet保存HTML5图像

我知道已经有很多关于StackOverflow的问题,但是我已经搜索了尽可能多的问题,并且还没有使我的代码正常工作,所以我终于发布了自己的问题.

我的目标是保存HTML5< canvas>在我的网页中找到服务器上的文件.我希望使用Java servlet来完成此任务.

我的JavasScript像这样捕获画布图像数据:

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尝试像这样保存图像:

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
}

该servlet成功写出了一个图像文件,但是无法正确打开,并且其中不包含所有图像数据.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]

有什么想法我在这里想念的吗?我觉得我犯了一些我无法发现的小错误.

解决方法:

通过使用multipart / form-data内容类型,具有相同的任务并能够使其工作(无需jQuery并借助maclema的答复):

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);

要异步进行,或者要发送的部分更多(例如,多张图片),或者要处理响应,请参见How to send multipart/form-data form content by ajax (no jquery)?

您的servlet的doPost方法如下所示:

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);
}

希望这可以帮助.

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