angularjs – 推送或取消进入$http(非全局)的transformRequest数组

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有关transFormRequest函数/数组的相当好的解释,请参阅 here的已接受答案.

在答案的最后一个例子中:

var transform = function(data){
    return $.param(data);
}

$http.post("/foo/bar",requestData,{
    headers: { 'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8'},transformRequest: transform
}).success(function(responseData) {
    //do stuff with response
});

但是,这个问题是transformRequest:transform会覆盖Angular预先构建的函数数组.

来自Angular docs

To globally augment or override the default transforms,modify the $httpProvider.defaults.transformRequest and $httpProvider.defaults.transformResponse properties. These properties are by default an array of transform functions,which allows you to push or unshift a new transformation function into the transformation chain. You can also decide to completely override any default transformations by assigning your transformation functions to these properties directly without the array wrapper. These defaults are again available on the $http factory at run-time,which may be useful if you have run-time services you wish to be involved in your transformations.

Similarly,to locally override the request/response transforms,augment the transformRequest and/or transformResponse properties of the configuration object passed into $http.

如果我想全局应用我的变换函数,我会这样做

$httpProvider.defaults.transformRequest.unshift(myFunction)

要么

$httpProvider.defaults.transformRequest.push(myFunction)

我的问题
如何将另一个转换函数推送到调用而不是全局?而不是擦除整个转换请求函数数组?

我找到了一个使用.concat方法的简单解决方
{
   transformRequest: [function(req){...}].concat($http.defaults.transformRequest)
}

或者,如果您希望在角度的默认转换后进行自定义转换.

{
   transformRequest: $http.defaults.transformRequest.concat([function(req){...}])
}

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