Question

我似乎无法弄清楚如何将多个源与knockoutjs的映射插件合并在一起。在下面的代码中，我尝试将不同的请求映射到相同的可观察集合中，但只添加了最后一个请求。

       this.dataSets = mapper.fromJS([], {
            key: (data) => data.dataSetKey,
            create: (options) => {
                console.log(options.data);
                var item = new FxsPortal.FxsSelectable<AscendDataSet>(
                    mapper.fromJS(options.data, {
                        'copy': ["cameraInfo"]
                    }));
                return item;
            }
        }); 

        userStore.get().subscriptions.forEach(subscription => {        
            this.logger.logInformation("Iterating Subscription", subscription);

            (xhr = $.ajax({
                beforeSend: (xhr) => xhr.setRequestHeader('Authorization', "Bearer " + OAuthModule.getStoredTokens().access_token),
                type: 'GET', url: AppOptions.DataServiceUri + "/" + subscription.id + "/datasets?includeCandidates=" + v,
            }))
                .done((result: { continuationToken; results }) => {

                    console.log(["Result from DataSet Api", result]);
                    var data = result.results;

                    mapper.fromJS(data, {}, this.dataSets);           
                }).always(() => {
                    this.isContentLoading(false);
                    xhr = null;
                });
        });

Answer 1

在调用ko.mapping之前合并数据。

jQuery.when可以等待多个延迟（ajax）对象。 http://api.jquery.com/jquery.when/

在这里，我们必须使用apply技巧将ajax个对象传递给$.when，因为我们正在处理未知的ajax调用计数。

//map your data array into an array of deferred objects, feed to $.when
$.when.apply($, $.map(userStore.get().subscriptions(), function(subscription) {
    return $.ajax(..); //without any 'done' 'always' 'fail' callbacks!
})).done(function() {
  //count of result is unknown, have to get them out from arguments
  $.each(arguments, function(i, arg) {
    // arg[0] is the JSON data, arg[1] and arg[2] are statusText, jqXHR.
  });

  // now, you got all JSON data, merge them before ko.mapping :-)
});

使用knockout.mapping插件将多个数组合并到一个可观察数组中

1 个答案: