13

I have an $.ajax promise and want to check whether my (syntactically valid) response contains an error, triggering the rejected status in that case.

I have worked with my own promise library which deals such tasks easily. I do not really like jQuery's Promise (cache) implementation with its Deferred object and may have overlooked something, because I seldom use it. I think the way to go is just using .then(), which seems to be rather complicated:

return $.ajax(...).then(function success(response) {
    var problem = hasError(response);
    if (problem) {
        var error = new $.Deferred;
        error.reject(problem);
        return error;
    } else
        return response;
});

This should return a promise which is rejected in case of network errors or problems with the response. But is returning a rejected deferred really the [only|best|shortest] way to go?

I also would appriciate help on how to deal with such "error-throwing response handlers" in the ajax options themselfes, I could not find good documentation about them.

Disclaimer: No, I cant change the server responses. The problem-detecting method is synchronous. I don't want to use other libraries, I'm particularly interested in the way jQuery solves this.

4

1 に答える 1

33

jQuery 1.8+ 用に更新されました

これに取り組む最も簡単な方法は、データの成功または失敗に基づいてフィルタリング$.ajaxするまでの応答を実行することです。.then

$.ajax()
    .then(function (response) {
        return $.Deferred(function (deferred) {
            var problem = hasError(response);

            if (problem) {
                return deferred.reject(problem)
            }

            deferred.resolve(response);
        }).promise();
    });

次に、この新しい promise を、これを消費する呼び出し元のコードに返すことができます。

var request = function () {
    return $.ajax()
        .then(function (response) {
            return $.Deferred(function (deferred) {
                var problem = hasError(response);

                if (problem) {
                    return deferred.reject(problem)
                }

                deferred.resolve(response);
            }).promise();
        });
};

request()
    .done(function (response) {
        // handle success
    })
    .fail(function (problem) {
        // handle failure
    });
于 2012-06-01T01:35:35.437 に答える