具有异步失败的 jake 任务不会导致失败通知冒泡

Question

使用以下代码，我希望任务“asyncfail”会导致一些错误输出进入控制台，并且退出代码不为零（就像任务“syncfail”一样）。但实际发生的情况是 jake 在没有进一步输出的情况下完成（下面的示例输出）并且进程退出代码指示成功。我不确定我是否对杰克做错了什么，或者承诺......有什么想法吗？

var Q = require("q");

task("syncfail", function() {
   fail("synchronous failure"); 
});

var http = require("http");

task("asyncfail", function() {

    waitForSleep().then(waitForSleep).then(waitForSleep).then(waitForSleep).then(function() {
        console.log("all done, time to fail");
        fail("This is the end");
    });

    console.log("finishing sync part");
}, { async:true});


function waitForSleep() {
  var deferred = Q.defer();

  setTimeout(function() {
    console.log("finished timeout");
    deferred.resolve();
  });

  return deferred.promise;
}

以下是任务 syncfail（按我预期完成）和 asyncfail（按我预期完成）的一些示例输出：

C:\src\letscodejavascript>node node_modules\jake\bin\cli.js -f .\jakerepro.js syncfail
jake aborted.
Error: synchronous failure
    at api.fail (C:\src\letscodejavascript\node_modules\jake\lib\api.js:221:18)
(See full trace by running task with --trace)

C:\src\letscodejavascript [master]> $lastexitcode
1

C:\src\letscodejavascript [master]> .\jake.bat -f .\jakerepro.js asyncfail

C:\src\letscodejavascript>node node_modules\jake\bin\cli.js -f .\jakerepro.js asyncfail
finishing sync part
finished timeout
finished timeout
finished timeout
finished timeout
all done, time to fail

C:\src\letscodejavascript [master]> $lastexitcode
0

Answer 1

fail() 不起作用的原因是 fail() 通过抛出异常以在外层捕获，但该异常被 .then() 的实现捕获。为了让 fail() 在以这种方式使用 Promise 时按预期工作，我需要在 setTimeout 中包装对 fail() 的调用，以便在 .then() 处理程序之外重新抛出错误。

var Q = require("q");

task("asyncfail", function() {

    waitForSleepPromise()
    .then(waitForSleepPromise)
    .then(waitForSleepPromise)
    .then(waitForSleepPromise)
    .then(function() {
        console.log("all done, time to fail");

        deferError(function() {
          fail("This is the end", "fasdf");
        });
    });

    console.log("finishing sync part");
}, { async:true});


function waitForSleepPromise() {

  var deferred = Q.defer();

  setTimeout(function() {
    console.log("finished timeout");
    deferred.resolve();
  }, 10);

  return deferred.promise;
}

function deferError(callback) {
  try {
    callback();
  }
  catch (err){
    setTimeout(function() {
      throw err;
    }, 0);
  }
}