如何通过AJAX POST正确传输data:image/jpeg;base64
数据网址。我现在有以下代码xhr.open('POST', 'http://url-sent-to/image/' + saveImage + '&imageid=' + imageid.value, true);
。
但是,网址http://url-sent-to/image/data:image/jpeg;base64,/9j/4AAQSkZJRgABAQAAAQABAAD…RRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFAH/2Q==&imageid=testimagedata
看起来不正确,尤其是因为其中包含=
。
$(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,作为我正在做的事情的一个例子。实质上,用户将选择图像,输入描述,然后点击渲染。当您检查Javascript控制台时,您将看到创建了一个Blob,并在底部看到了POST消息:http://jsfiddle.net/mattography/Lgduvce1/2/
答案 0 :(得分:0)
您正在寻找encodeURI()
,这将完全符合您的需求。
请注意,您错过了?
来启动查询字符串。
另请注意,制作长篇网址不是一个好主意;你应该发送一个POST请求。