URL编码"data: image /jpg；base64“图片

Question

如何对data:image/jpeg;base64数据url进行编码，以便通过AJAX POST正确传输。我现在有下面的代码xhr.open('POST', 'http://url-sent-to/image/' + saveImage + '&imageid=' + imageid.value, true);可以做到这一点。

但是，URL http://url-sent-to/image/data:image/jpeg;base64,/9j/4AAQSkZJRgABAQAAAQABAAD…RRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFAH/2Q==&imageid=testimagedata看起来并不正确，特别是因为它包含=。

    $(function () {
  var fileInput = document.getElementById("file")
    , renderButton = $("#renderButton")
    , imgly = new ImglyKit({
        container: "#container",
        ratio: 1 / 1
      });

  // As soon as the user selects a file...
  fileInput.addEventListener("change", function (event) {
    var file;

    var fileToBlob = event.target.files[0];
          var blob = new Blob([fileToBlob], {"type":fileToBlob.type});
          // do stuff with blob
          console.log(blob);
    // Find the selected file
    if(event.target.files) {
      file = event.target.files[0];
    } else {
      file = event.target.value;
    }

    // Use FileReader to turn the selected
    // file into a data url. ImglyKit needs
    // a data url or an image
    var reader = new FileReader();
    reader.onload = (function(file) {
      return function (e) {
        data = e.target.result;

        // Run ImglyKit with the selected file
        try {
          imgly.run(data);
        } catch (e) {
          if(e.name == "NoSupportError") {
            alert("Your browser does not support canvas.");
          } else if(e.name == "InvalidError") {
            alert("The given file is not an image");
          }
        }
      };
    })(file);
    reader.readAsDataURL(file);
  });

  // As soon as the user clicks the render button...
  // Listen for "Render final image" click
  renderButton.click(function (event) {
    var dataUrl;


    imgly.renderToDataURL("image/jpeg", { size: "1200" }, function (err, dataUrl) {
      // `dataUrl` now contains a resized rendered image with
      // a width of 300 pixels while keeping the ratio

       //Convert DataURL to Blob to send over Ajax
        function dataURItoBlob(dataUrl) {
        // convert base64 to raw binary data held in a string
        // doesn't handle URLEncoded DataURIs - see SO answer #6850276 for code that does this
        var byteString = atob(dataUrl.split(',')[1]);

        // separate out the mime component
        var mimeString = dataUrl.split(',')[0].split(':')[1].split(';')[0];

        // write the bytes of the string to an ArrayBuffer
        var ab = new ArrayBuffer(byteString.length);
        var ia = new Uint8Array(ab);
        for (var i = 0; i < byteString.length; i++) {
            ia[i] = byteString.charCodeAt(i);
        }
        // write the ArrayBuffer to a blob, and you're done
        //var bb = new BlobBuilder();
        //bb.append(ab);
        //return bb.getBlob(mimeString);
    }


    var blob = dataURItoBlob(dataUrl);
    var fd = new FormData(document.forms[0]);
    var xhr = new XMLHttpRequest();

    var saveImage = dataUrl;
    //console.log(saveImage);


    fd.append("myFile", blob);
    xhr.open('POST', 'http://url-sent-to/image/' + saveImage + '&imageid=' + imageid.value, true);
    xhr.send(fd);

我有一个小提琴设置，作为我正在做的一个例子。本质上，用户将选择一个图像，输入一个描述，然后点击render。当您查看Javascript控制台时，您将看到创建了一个Blob，并在底部显示POST消息：http://jsfiddle.net/mattography/Lgduvce1/2/

Answer 1

您正在寻找的是encodeURI()，它将执行您正在寻找的任务。

请注意，查询字符串的开头缺少一个?。

另请注意，使URL过长并不是一个好主意；您应该改为发送POST请求。