toDataURL的输出是base64，如何减少1/3的上传时间和带宽？ [英] toDataURL's output is base64, how to reduce the uploading time and bandwidth of a factor 1/3?

问题描述

以下代码的目标是压缩输入文件（2 MB JPG文件=> 500 KB文件），然后在提交< form> 。

The goal of the following code is to compress the input file (2 MB JPG file => 500 KB file) and then upload it to server when submitting the <form>.

将图像从JPG文件导入画布，然后使用 toDataURL with：

When importing an image from a JPG file into a canvas, and exporting it with toDataURL with:

function doit() {
    var file = document.getElementById('file').files[0],
        canvas = document.getElementById('canvas'),
        hidden = document.getElementById('hidden'),
        ctx = canvas.getContext("2d"),
        img = document.createElement("img"),
        reader = new FileReader();  
    
    reader.onload = function(e) { 
        img.src = e.target.result;
    }
    
    img.onload = function () { 
        ctx.drawImage(img, 0, 0);
        hidden.value = canvas.toDataURL("image/jpeg", 0.5);
    }
    reader.readAsDataURL(file);
}

<input type="file" onchange="doit();" id="file" />

<form action="/upload" method="post">
<input type="hidden" id="hidden" />
<input type="submit" />
</form>

<canvas id="canvas" style="display: none" />

它有效，但< form> 隐藏字段>是base64编码的，例如：

it works, but the output hidden field in the <form> is base64-encoded, i.e. something like:

data:image/jpeg;base64,/9j/4AAQSkZJRgAB...

众所周知，base64使用正常大小的二进制数据的1.3333倍。

It is well known that base64 uses 1.3333 times the normal size of binary data.

问题：在这种情况下如何避免浪费上传时间（客户端=>服务器）和带宽的1/3数据，即提交<时; form> ？

Question: how to avoid to waste 1/3 of data in uploading time (client => server) and bandwidth in this case, i.e. when submitting the <form>?

注意：如果我使用AJAX而不是<$ c $，我认为问题会是一样的c>< form> 提交，不是吗？

Note: I think the problem will be the same if I use AJAX instead of <form> submission, isn't it?

推荐答案

var jsForm = null;

function doit() {
    var file = document.getElementById('file').files[0],
        canvas = document.getElementById('canvas'),
        ctx = canvas.getContext("2d"),
        img = document.createElement("img");

    img.src = window.URL.createObjectURL(file);
    
    img.onload = function () {
        if (!jsForm) {
          jsForm = new FormData();
        }
        ctx.drawImage(img, 0, 0);
        canvas.toBlob(function(blob) {
          jsForm.set('image', blob, file.name);
        }, "image/jpeg", 0.5);
    }
}

var form = document.getElementById('form');
form.onsubmit = function(e) {
  e.preventDefault();
  if (!jsForm) return;
  var request = new XMLHttpRequest();
  request.open(this.method||'POST', this.action||'/');
  request.send(jsForm);
  jsForm = null;
}

<form method="POST" action ="/upload" id="form">
  <input type="file" onchange="doit();" id="file" />
  <button>Submit</button>
</form>
<canvas id="canvas" style="display: none" />

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