帶異步故障的傑克任務不會導致故障通知冒泡

Question

使用以下代碼，我期望任務「asyncfail」導致某些錯誤輸出轉到控制檯並且退出代碼爲非零（就像任務「syncfail」一樣）。實際發生的是jake完成後沒有進一步輸出（下面的示例輸出），並且進程退出代碼表示成功。我不確定我是否在做傑克錯誤或承諾...任何想法？

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; 
} 

下面是任務syncfail（其中完成，我會期望）和asyncfail一些示例輸出（其中沒有完成，我會期望）：

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

原因失敗（）沒有工作是失敗（）通過拋出一個異常被捕獲到一個外層，但是這個異常被實現.then（）所捕獲。爲了使fail（）以這種方式按照預期工作，我需要在setTimeout中調用fail（）來重新拋出.then（）處理程序之外的錯誤。

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); 
    } 
}