AngularJS 函数返回带有所需参数的 Promise

Question

简单的问题。我已经构建了这个功能:

// Gets the kit by id or slug
var _getKit = function (id) {

    // Try to get our kit using our shared service
    return sharedKitService.get(id).then(function (response) {

        // Assign the response to our service
        service.models.kit = response;

        // Return our response
        return response;
    })
};

我想向该函数添加一个检查，如下所示:

// Gets the kit by id or slug
var _getKit = function (id) {

    // If we have no id, exit the function
    if (!id)
        return;

    // Try to get our kit using our shared service
    return sharedKitService.get(id).then(function (response) {

        // Assign the response to our service
        service.models.kit = response;

        // Return our response
        return response;
    })
};

但我知道这行不通，因为如果没有 id 那么该函数将不再产生 promise 。我知道我可以做这样的事情:

// Gets the kit by id or slug
var _getKit = function (id) {

    // If we have no id
    if (!id) {

        // Defer our promise
        var deferred = $q.derfer();

        // Reject our promise
        deferred.reject();

        // Return our promise
        return deferred.promise;
    }

    // Try to get our kit using our shared service
    return sharedKitService.get(id).then(function (response) {

        // Assign the response to our service
        service.models.kit = response;

        // Return our response
        return response;
    })
};

但这似乎有点矫枉过正。有没有更简单的方法来做到这一点？

Answer 1

来自Angular docs :

只需返回$q.reject(reason)。这将返回 Promise，该 Promise 会立即被拒绝