Question

我编写了一个函数，通过递归搜索对象来获取所有特定元素，并返回结果元素数组。

该功能本身运作完美，但由于某种原因，＆＃34;返回＆＃34; part不会被赋值给调用函数的变量。

这是函数本身：

var getLinksInObject = function(object){
    // Define Variables
    var allLinks = [];
    var getLinksTimeout = null;

    // Function to get the links in an object. Called recursively
    var getLinks = function(object){
        // Get category and link links
        var categories = object.categories;
        var links = object.links;

        // If there are links
        if (links && links.length > 0){
            // Push each of them to the allLinks array
            for (var i=0, j=links.length; i<j; i++){
                allLinks.push(links[i]);
            };
        };

        // If there are category links, push them to the allLinks array
        if (categories && categories.length > 0){
            for (var i=0, j=categories.length; i<j; i++){
                // Build the link
                var link = {
                    title: categories[i].title,
                    link: categories[i].link
                };
                // Push the link to allLinks
                allLinks.push(link);

                // If there are sub-links, run getLinks on that object to get the next level of links
                if (categories[i].categories) {
                    // Reset the listener so it knows we are going for another round
                    resetTimeout();

                    // Get the links in the next level
                    getLinks(categories[i]);
                }           
            };
        };
    };

    // Listens for when the recursive calls finish and returns the result
    var resetTimeout = function(){
        clearTimeout(getLinksTimeout);
        getLinksTimeout = setTimeout(function(){
            log(allLinks);
            return allLinks;
        }, 50);
    };

    resetTimeout();
    getLinks(object);
};

这是一个运行它的测试对象，它匹配实际对象的结构：

{
    link: "http://test.com",
    title: "This is a test",
    categories: [
        {
            link: "http://test.com",
            title: "This is a test",
            categories: [
                {
                    link: "http://test.com",
                    title: "This is a test",
                    categories: [
                        {

                        }
                    ]
                }
            ]
        },
        {
            link: "http://test.com",
            title: "This is a test",
            categories: [
                {
                    link: "http://test.com",
                    title: "This is a test",
                    categories: [
                        {
                            link: "http://test.com",
                            title: "This is a test",
                            categories: [
                                {
                                    link: "http://test.com",
                                    title: "This is a test"
                                }
                            ]
                        }
                    ]
                },
                {
                    link: "http://test.com",
                    title: "This is a test",
                    categories: [
                        {

                        }
                    ]
                },
                {
                    link: "http://test.com",
                    title: "This is a test",
                    categories: [
                        {

                        }
                    ]
                }
            ]
        },
        {
            link: "http://test.com",
            title: "This is a test",
            categories: [
                {
                    link: "http://test.com",
                    title: "This is a test",
                    categories: [
                        {

                        }
                    ]
                }
            ]
        }
    ]
}

这是一个重复问题的小提琴。您将能够在控制台中看到问题，因为记录了＆＃34;结果＆＃34;以未定义的方式注销。

http://jsfiddle.net/KS7LG/

Answer 1

您分配给getLinksInObject的功能没有return声明。它将始终返回undefined。

整个代码中唯一的return语句是：

setTimeout(function(){
console.log(allLinks);
    return allLinks;
}, 50);

setTimeout正在调用该函数，setTimeout不会关注它调用的函数的任何返回值。

setTimeout是异步的，它稍后运行代码并且不会阻止从继续调用它的函数。如果你想处理在时间用完之后生成的数据，那么你必须从你传递的函数开始前进，现在回传数据为时已晚。

Answer 2

Javascript（主要）同步运行，因此您不需要任何timeout相关代码来使function正常工作。在计算时立即返回值就可以了。

请参阅：jsFiddle

特别是，以下变化：

              //resetTimeout(); <- remove this

                // Get the links in the next level
                getLinks(categories[i]);
            }           
        };
    };
};

//removed timeout code

getLinks(object);
return allLinks;

Answer 3

返回将无法与setTimeout一起使用，因为它是异步的。您需要创建一个回调来实现您想要的目标：

var getLinksInObject = function(object, callback) {
    // ...            
    var resetTimeout = function(){
        clearTimeout(getLinksTimeout);
        getLinksTimeout = setTimeout(function(){
            log(allLinks);
            callback(allLinks);
        }, 50);
    };

   // ...
};

然后您必须修改getLinksInObject电话：

getLinksInObject(function(allLinks) {
     // do what you need to do with allLinks here
});

函数“返回”未分配给变量

3 个答案: