$ Q承诺不解决 [英] $q promise not resolving

问题描述

我想不通这是为什么不解决，任何想法？ 解决这个可以打印，但它永远不会使它返回到。然后承诺的分辨率。

  VAR承诺= wait（）方法;
promise.then（功能（结果）{
    的console.log（等待返回 - +结果）;
}）;功能等待（）{
    变种推迟= $ q.defer（）;
    如果（忙）{
        的setTimeout（函数（）{
            等待（）;
        }，500）;
    }其他{
        的console.log（解决此）;
        deferred.resolve（等待已经结束了。）;
    }
返回deferred.promise;
};
 

解决方案

下面是如何可以代替完成：

  VAR承诺= wait（）方法;
promise.then（功能（结果）{
    的console.log（等待返回 - +结果）;
}）;功能等待（）{
  变种推迟= $ q.defer（）;
  （函数_wait（）{
    如果（忙）{
      的setTimeout（_wait，500）;
    }其他{
      的console.log（解决此）;
      deferred.resolve（等待已经结束了。）;
    }
  }）（）;
  返回deferred.promise;
};
 

关键区别在于，将只有一个延迟，创建一个包装函数返回。这种延迟将 _wait 函数来最终解决。

在你的情况，以后每次（递归）等待（）调用创建一个不同的延迟对象。其中一个对象将最终解决，正确的 - 但是，这将是由第一个返回的同一个对象等待（）调用只有当忙将在这一刻假的。显然，大多数时候也不会

I can't figure out why this isn't resolving, any ideas? "resolve this" does print, but it never makes it back to the resolution of the promise in .then.

var promise = wait(); 
promise.then(function(result){
    console.log("wait returned - " + result); 
}); 

function wait(){
    var deferred = $q.defer();
    if(busy){ 
        setTimeout(function(){
            wait(); 
        },500); 
    } else {
        console.log("resolve this");
        deferred.resolve("Wait is over."); 
    }
return deferred.promise; 
}; 

解决方案

Here's how it can be done instead:

var promise = wait(); 
promise.then(function(result){
    console.log("wait returned - " + result); 
}); 

function wait(){
  var deferred = $q.defer();
  (function _wait() {
    if (busy) { 
      setTimeout(_wait, 500);
    } else {
      console.log("resolve this");
      deferred.resolve("Wait is over."); 
    }
  })();
  return deferred.promise;
};

The key difference is that there will be only one deferred, created and returned by a 'wrapper' function. This deferred will be eventually resolved by _wait function.

In your case, each subsequent (recursive) wait() call creates a different deferred object. One of these objects will be eventually resolved, right - but that will be the same object as returned by the first wait() call only if busy will be false at this moment. Apparently, most of the time it won't.

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