javascript - AngularJS : $scope. $watch 一个对象并只返回它改变的属性

Question

我正在做一些简单的范围观察，例如:

$scope.$watch('myObject', function(newValue, oldValue) {
    if (newValue !== oldValue) {
       return newValue;
    }
}, true);

其中 myObject 是具有多个属性的普通对象。我只想返回更改的属性，即，如果有一个属性发生更改，比如 myObject.changedProperty，我只想返回它。

但我想监视整个对象(因此，我不必为每个属性设置不同的监视)。如何才能做到这一点？

谢谢!

Answer 1

$watchCollection 做你想做的。 ( $rootScope.Scope )

$watchCollection(obj, listener);

Shallow watches the properties of an object and fires whenever any of the properties change (for arrays, this implies watching the array items; for object maps, this implies watching the properties). If a change is detected, the listener callback is fired.

The obj collection is observed via standard $watch operation and is examined on every call to $digest() to see if any items have been added, removed, or moved. The listener is called whenever anything within the obj has changed. Examples include adding, removing, and moving items belonging to an object or array.