$ .getJSON的外部访问JSON数据（） [英] Access json data outside of $.getJSON()

问题描述

$(document).ready(function () {
    var value = getParmsVals()["search"];
    $.getJSON('/api/search/GetQuestionByKey/' + value, function (jsonData) {
        $(jsonData).each(function (i, item) {
            var name = getAuthorName(item.userId);
        });
    });
});

function getAuthorName(userId) {
    var fullname = "default";
    $.getJSON('/api/search/GetUserById/' + userId, function (jsonData) {
        fullname = jsonData.firstname + " " + jsonData.lastname;
    });
    return fullname;
}

我试图通过调用getAuthorName方法来访问全名变量，但我无法得到正确的值。它总是给我的价值默认。

I'm trying to access the fullname variable by calling the getAuthorName method but I couldn't get the correct value. It's always giving me the value "default".

推荐答案

您不会收益从异步方法，你可以看到，这是行不通的！你需要的是一个回调函数，考虑：

You wouldn't return from an async method, as you can see, it doesn't work! What you need is a callback function, consider:

function getAuthorName(userId, callback) {
    var fullname = "default";
    $.getJSON('/api/search/GetUserById/' + userId, function (jsonData) {
        fullname = jsonData.firstname + " " + jsonData.lastname;
        callback(fullname);
    });
}

注意我们是如何通过在回调，然后在你的get调用月底打电话吗？现在，调用此方法像这样：

Notice how we pass in callback and then call it at the end of your get call? Now call this method like so:

getAuthorName(userID, function(name) {
    console.log(name);
});

现在，你有机会获得全名在回调函数！

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