Question

我正在尝试使它成为document.write（）9，然后暂停1秒，写8，然后暂停1秒，写7，然后暂停1秒，依此类推，以此类推，直到它命中0。但是现在它只是一次打印出整个内容。我在做什么错了？

var delay = function(){
setTimeout(delay, 1000); // check again in a second
}

var count = 10;
while (count > 0) {
count--;
delay();
document.write(count);
}

9876543210

Answer 1

两个问题：

调用setTimeout仅使函数排队等待超时后执行-不会暂停主线程。
document.write是document.wrong，通常-如果文档已经被解析，它将被新的HTML字符串替换。请改用正确的DOM方法。

您可以await每秒钟承诺并附加span项：

const delay = () => new Promise(res => setTimeout(res, 1000));
(async () => {
  let count = 10;
  while (count > 0) {
    await delay();
    count--;
    document.body.appendChild(document.createElement('span')).textContent = count;
  }
})();

您还可以（同步）设置每次迭代的超时时间：

for (let count = 9; count >= 0; count--) {
  setTimeout(() => {
    document.body.appendChild(document.createElement('span')).textContent = count;
  }, (10 - count) * 1000);
}

Answer 2

setTimeout是异步的，它不会暂停执行。

var count = 10;
var delay = function(){
  if (count > 0) {
    count--;
    document.write(count);
    setTimeout(delay, 1000); // check again in a second
  }
}
delay();

这可能有效，但是在页面加载完成后，我认为document.write不起作用。这意味着您无法在超时回调中调用它。

Answer 3

下面是使用searchterms = ['cheese', 'hippos', 'whales', 'beluga'] async def sendQueries(queries, deposit=list()): with concurrent.futures.ThreadPoolExecutor(max_workers=4) as executor: loop = asyncio.get_event_loop() futures = [ loop.run_in_executor( executor, searching(queries) ) ] for response in await asyncio.gather(*futures): deposit.append(response.json()) return deposit def running(): loop = asyncio.get_event_loop() loop.run_until_complete(loop.create_task(sendQueries(searchterms))) loop.close() print(running()) print(str(time.time() - x))和setInterval的示例

console.log

正确使用setTimeout

3 个答案: