How can you access the URL of a resource action in Angular?

In my Angular application, I have created a resource named 'Files' with the following definition:

app.factory('Files', function($resource) {
    return $resource('/api/accounts/:account_id/sites/:site_id/files/:file_id');
});

If I want to create a new file, I use the following syntax:

var file = new Files($scope.file);
file.$save( $stateParams );

Now, I am wondering how I can retrieve the URL that is used by file.$save (POST /api/accounts/:account_id/sites/:site_id/files) in my controller, where all $stateParams parameters are already filled in.

UPDATE:

I am currently working on uploading a file using ng-file-upload. This module requires the URL to be specified when calling the upload function:

$scope.upload = function (file) {
    var url = 'api/accounts/' + $stateParams.account_id + '/sites/' + $stateParams.site_id + '/files';
    Upload.upload({
        url: url,
        data: {file: file}
    }).then(function (resp) {
        console.log('Success ' + resp.config.data.file.name + 'uploaded. Response: ' + resp.data);
        getFiles();
    }, function (resp) {
        console.log('Error status: ' + resp.status);
    }, function (evt) {
        var progressPercentage = parseInt(100.0 * evt.loaded / evt.total);
        console.log('progress: ' + progressPercentage + '% ' + evt.config.data.file.name);
    });
};

I am trying to find a way to avoid manually constructing the URL variable. After looking at the angular $resource source code, it seems this might not be possible. Is there a workaround for this?

Answer №1

To achieve your goal, consider refactoring your code in the following way:

app.factory('fileUrl', ['accountId', 'siteId', 'fileId', function fileUrlFactory(accountId, siteId, fileId) {
  return '/api/accounts/' + accountId + '/sites/' + siteId + '/files/' + fileId';
}]);

By using this factory, you can centralize the construction of file URLs in your code. Then use it like this:

var yourFileUrl = fileUrl($stateParams.account_id, 22, 41); // Or any other necessary parameters
var file = $resource(yourFileUrl);
file.$save();

Remember that $resource acts as a factory itself, allowing you to create instances with specific parameters.

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