Javascript-等待数组完成填充
我试图得到一些数组元素的值。它适用于元素[0]、[1]、[2]、[3],但不适用于元素[4]Javascript-等待数组完成填充,javascript,arrays,asynchronous,Javascript,Arrays,Asynchronous,我试图得到一些数组元素的值。它适用于元素[0]、[1]、[2]、[3],但不适用于元素[4] function getBase64() { const urls = ['https://i.imgur.com/egNg7JU.jpg', 'https://i.imgur.com/RLZ7WH1.jpg', 'https://i.imgur.com/qfabBbA.jpg', 'https://i.imgur.com/Zuh1KaX.jpg', 'https://i.imgur
function getBase64() {
const urls = ['https://i.imgur.com/egNg7JU.jpg',
'https://i.imgur.com/RLZ7WH1.jpg', 'https://i.imgur.com/qfabBbA.jpg',
'https://i.imgur.com/Zuh1KaX.jpg', 'https://i.imgur.com/yD7X6Q1.jpg'
];
let base64urls = [];
const start = async () => {
await asyncForEach(urls, async (num) => {
await waitFor(50)
toDataURL(num, function(dataURL) {
base64urls.push(dataURL);
});
})
console.log(base64urls);
console.log(base64urls[4]);
}
start()
}
async function asyncForEach(array, callback) {
for (let index = 0; index < array.length; index++) {
await callback(array[index], index, array)
}
}
const waitFor = (ms) => new Promise(r => setTimeout(r, ms))
它看起来像是
toDataURL
是异步的,并且是基于回调的-要么更改它,使其返回Promise
和wait
该Promise
,要么将Promise
的解析
传递到回调中:
async function getBase64() {
const urls = ['https://i.imgur.com/egNg7JU.jpg',
'https://i.imgur.com/RLZ7WH1.jpg', 'https://i.imgur.com/qfabBbA.jpg',
'https://i.imgur.com/Zuh1KaX.jpg', 'https://i.imgur.com/yD7X6Q1.jpg'];
const base64urls = [];
for (const url of urls) {
const dataURL = await new Promise(resolve => toDataURL(url, resolve));
base64urls.push(dataURL);
}
console.log(base64urls);
console.log(base64urls[4]);
}
function toDataURL(src) {
return new Promise(resolve => {
const image = new Image();
image.crossOrigin = 'Anonymous';
image.onload = function () {
const canvas = document.createElement('canvas');
const context = canvas.getContext('2d');
canvas.height = this.naturalHeight;
canvas.width = this.naturalWidth;
context.drawImage(this, 0, 0);
const dataURL = canvas.toDataURL('image/jpeg');
resolve(dataURL);
};
image.src = src;
});
}
如果要将toDataURL
函数更改为返回承诺,以便不必将其视为回调:
async function getBase64() {
const urls = ['https://i.imgur.com/egNg7JU.jpg',
'https://i.imgur.com/RLZ7WH1.jpg', 'https://i.imgur.com/qfabBbA.jpg',
'https://i.imgur.com/Zuh1KaX.jpg', 'https://i.imgur.com/yD7X6Q1.jpg'];
const base64urls = [];
for (const url of urls) {
const dataURL = await new Promise(resolve => toDataURL(url, resolve));
base64urls.push(dataURL);
}
console.log(base64urls);
console.log(base64urls[4]);
}
function toDataURL(src) {
return new Promise(resolve => {
const image = new Image();
image.crossOrigin = 'Anonymous';
image.onload = function () {
const canvas = document.createElement('canvas');
const context = canvas.getContext('2d');
canvas.height = this.naturalHeight;
canvas.width = this.naturalWidth;
context.drawImage(this, 0, 0);
const dataURL = canvas.toDataURL('image/jpeg');
resolve(dataURL);
};
image.src = src;
});
}
然后const-dataURL=wait-toDataURL(url)
在这种情况下,您可以使用它来等待查询结果
constURL=['https://i.imgur.com/egNg7JU.jpg',
'https://i.imgur.com/RLZ7WH1.jpg', 'https://i.imgur.com/qfabBbA.jpg',
'https://i.imgur.com/Zuh1KaX.jpg', 'https://i.imgur.com/yD7X6Q1.jpg'];
让base64url=[];
Promise.all(url.map(url=>fetch(url)))。然后(res=>toBase64DataURL(res))。然后(result=>{base64url.push(result.toDataURL());
log(base64url);});
函数toBase64DataURL(src){
返回新承诺(解决=>{
常量图像=新图像();
image.crossOrigin='匿名';
image.onload=\u=>{
const canvas=document.createElement('canvas');
const context=canvas.getContext('2d');
canvas.height=this.naturalHeight;
canvas.width=this.naturalWidth;
drawImage(this,0,0);
const dataURL=canvas.toDataURL('image/jpeg');
解析(dataURL);
};
image.src=src;
});
}
可能重复的等待等待(50)
为什么会出现小停顿?听起来您并不是在等待异步操作完成,而是交叉手指,希望您已经等待了很长时间。对不起,我到底应该在我的toDataURL
中更改什么?我已经将它添加到我的问题中。我发布的代码片段将一个承诺解析传递到基于回调的toDataURL
,允许它被等待
,因此您不需要用此更改toDataURL
。我尝试了第一种方法,但现在整个数组是空的:/