实施node.js的回调超时

问题描述：

这是在node.js中的典型情况：

This is a typical situation in node.js:

asyncFunction(arguments, callback);

在 asynFunction 完成后，回调被调用。我用这种模式看到的一个问题是，如果 asyncFunction 的从不的结束（和 asynFunction 没有内置超时系统），那么回调将永远不会被调用。更糟的是，它似乎回调没有确定 asynFunction 将永远不会返回的路。

When asynFunction completes, callback gets called. A problem I see with this pattern is that, if asyncFunction never completes (and asynFunction doesn't have a built-in time-out system) then callback will never be called. Worse, it seems that callback has no way of determining that asynFunction will never return.

我想实现一个超时，即如果回调未在1秒钟之内被称为由 asyncFunction ，那么回调自动被调用，并假设 asynFunction 已经出错了。什么是这样做的标准方式？

I want to implement a "timeout" whereby if callback has not been called by asyncFunction within 1 second, then callback automatically gets called with the assumption that asynFunction has errored out. What is the standard way of doing this?

答

我不熟悉做任何库，但它不是很难线了自己。

I'm not familiar with any libraries that do this, but it's not hard to wire up yourself.

// Setup the timeout handler
var timeoutProtect = setTimeout(function() {

  // Clear the local timer variable, indicating the timeout has been triggered.
  timeoutProtect = null;

  // Execute the callback with an error argument.
  callback({error:'async timed out'});

}, 5000);

// Call the async function
asyncFunction(arguments, function() {

  // Proceed only if the timeout handler has not yet fired.
  if (timeoutProtect) {

    // Clear the scheduled timeout handler
    clearTimeout(timeoutProtect);

    // Run the real callback.
    callback();
  }
});

实施node.js的回调超时

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