javascript - 将 AJAX GET 响应的一部分(JSON 格式)存储到字符串变量中时出现问题

Question

我在尝试通过 AJAX 将 GET 请求的部分结果存储到字符串变量中时遇到问题。

基本上，我想让包含 GET 请求操作的某个函数返回该操作的结果。

var count = 0;

$.getJSON("https://api.icndb.com/jokes/count", function(data){ 
    count = data.value+1;
    for (i = 1; i < count; i++){ 
        if (i != 1) {
            setTimeout(jokeGet, i*7500, i);
        }
        else {
            jokeGet(i);
        }
    }
});

function jokeGet(n) {
    var str = "";
    $.getJSON("https://api.icndb.com/jokes/" + n, function(data){
        if (data.type != "NoSuchQuoteException") {
            $(".joke").html(data.value.joke);
            str = data.value.joke;
        }
        else {
            count++;
        }
    });

    return str;
}

我发出请求的 API 将信息存储在 JSON 树中。以下是此类树的两个示例:

{ "type": "success", "value": { "id": 1, "joke": "Chuck Norris uses ribbed condoms inside out, so he gets the pleasure.", "categories": ["explicit"] } }

{ "type": "NoSuchQuoteException", "value": "No quote with id=8." }

但是，每当我运行单元测试(通过 QUnit)时，结果表明，在任何情况下，jokeGet() 函数都会返回一个空字符串。这让我觉得很奇怪，因为我认为 str = data.value.joke 行会做到这一点，因此笑话存储在该变量 str 中。

显然，因为 str 总是返回空字符串，但事实并非如此。关于这是为什么的任何建议？

更新

考虑到我现在所做的目标不是让程序运行，而是进行单元测试以证明程序运行，我决定包含“单元测试”文件:

QUnit.test("cn_jokes", function(assert) {
    function joke(n, expected) {
        assert.equal(jokeGet(n), expected);
    }
    joke(1, "Chuck Norris uses ribbed condoms inside out, so he gets the pleasure.");
    joke(8, undefined);
    joke(163, "Ninjas want to grow up to be just like Chuck Norris. But usually they grow up just to be killed by Chuck Norris.");
    joke(221, "Chuck Norris is the only person to ever win a staring contest against Ray Charles and Stevie Wonder.");
    joke(352, "Chuck Norris doesn't see dead people. He makes people dead.");
    joke(502, "Chuck Norris insists on strongly-typed programming languages.");
    joke(526, "No one has ever pair-programmed with Chuck Norris and lived to tell about it.");
    joke(598, "Once Chuck Norris and Superman had a competition. The loser had to wear his underwear over his pants.");
});

如您所见，我想要获取 jokeGet() 函数，特别是返回笑话值。请告诉我这是否可行。

Answer 1

$.getJSON 是异步的；当发出请求时，您的代码将继续运行。因此，str 早在您传递给 getJSON 的回调运行之前就返回了。您可能应该让 jokeGet 接受一个回调函数，并在请求完成时调用它(将 data.value.joke 作为参数传递):

function jokeGet(n, callback) {
    $.getJSON("https://api.icndb.com/jokes/" + n, function(data){
        if (data.type != "NoSuchQuoteException") {
            $(".joke").html(data.value.joke);
            if (callback !== undefined && callback !== null)
                callback(data.value.joke);
        }
        else {
            count++;
            if (callback !== undefined && callback !== null)
                callback(undefined); // not sure if you want undefined or "" in this case
        }
    });
}

编辑:您可以在 QUnit 中使用异步回调。只需使用 assert.async()，如前所述 here :

QUnit.test("cn_jokes", function(assert) {
    var done = assert.async();
    var jokesDone = 0;
    var numJokes = 8; // make sure to change this if you add more
    function joke(n, expected) {
        jokeGet(n, function(j) {
            assert.equal(j, expected);
            if (++jokesDone == numJokes) done();
        }
    }
    joke(1, "Chuck Norris uses ribbed condoms inside out, so he gets the pleasure.");
    joke(8, undefined);
    joke(163, "Ninjas want to grow up to be just like Chuck Norris. But usually they grow up just to be killed by Chuck Norris.");
    joke(221, "Chuck Norris is the only person to ever win a staring contest against Ray Charles and Stevie Wonder.");
    joke(352, "Chuck Norris doesn't see dead people. He makes people dead.");
    joke(502, "Chuck Norris insists on strongly-typed programming languages.");
    joke(526, "No one has ever pair-programmed with Chuck Norris and lived to tell about it.");
    joke(598, "Once Chuck Norris and Superman had a competition. The loser had to wear his underwear over his pants.");
});