我有一个图像裁剪器,它可以接收图像,并立即将其转换为blob
Blob {type: "image/jpeg", size: 40092, slice: function}size: 40092type:
"image/jpeg"__proto__: Blob
在加载到画布之前。裁剪后,图像以base64输出。我需要能够将base64图像转换回blob或图像,这样我就可以通过AJAX将图像POST
到API端点。我有以下内容可以完成所有这些,但我不知道如何将渲染的图像转换回POST
的blob。用于POST
的当前URL在控制台中使用encodeURIComponent(dataUrl)
:点击render后如下所示
http://url.com/rest/v1/utils/guid/encode?data%3Aimage%2Fjpeg%3Bbase64%2C%2F…hLGoZInTrWlH2qG7Az0okBRUnvTqXsaSoAKKKK0jsADrTqaOtOqgP%2F9k%3D&imageid=Test
这是底部小提琴中使用的代码
$(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 = encodeURIComponent(dataUrl);
//console.log(saveImage);
fd.append("myFile", blob);
xhr.open('POST', 'http://url.com/rest/v1/utils/guid/encode?' + saveImage + '&imageid=' + imageid.value, true);
xhr.send(fd);
var image = $("<img><br>").attr({
src: dataUrl
});
image.appendTo($(".result"))
$button = $('<button class="btn btn-default remove">')
.text('Remove')
.on('click', function () {
image.remove();
$(this).remove();
return false;
});
$button.appendTo($(".result"));
});
});
});
JSFIDDLE:JSFIDDLE
dataURItoBlob
没有返回任何内容,它应该返回一个blob
return new Blob([ab], {type: 'image/jpeg'});
此外,为什么要将图像的数据url放在ajax post请求的url中?