【问题标题】:Linkedin People Search Javascript API not returning any resultLinkedin People Search Javascript API 未返回任何结果
【发布时间】:2011-11-17 07:51:15
【问题描述】:

我正在使用linkedin JS API

我正在尝试执行people search,它没有返回任何内容

代码:

<script type="text/javascript" src="http://platform.linkedin.com/in.js">
        api_key: --my api key--
        onLoad: onLinkedInLoad
        authorize: true
</script> 

<script type="text/javascript">

        function onLinkedInLoad() {
            IN.Event.on(IN, "auth", onLinkedInAuth);
        }

        function onLinkedInAuth() {

            IN.API.PeopleSearch()
            .fields("firstName", "lastName", "distance")
            .params({"company-name":"infosys"})
            .result(displayPeopleSearch)
            .error(displayPeopleSearchError);

        }

        function displayPeopleSearch(){

            var peopleSearchDiv = document.getElementById("peoplesearch");

            var members = peopleSearch.people.values;


            for (var member in members) {
            // but inside the loop, everything is the same
            // extract the title from the members first position
                peopleSearchDiv.innerHTML += "<p>" + members[member].firstName + " " + members[member].lastName + " is a " + members[member].positions.values[0].title + ".</p>";

            }
        }

任何帮助,我做错了什么?

我正在关注这个tutorial

【问题讨论】:

    标签: javascript api search linkedin


    【解决方案1】:

    在这一行中,您使用了一个未在任何地方声明的变量 peopleSearch

    var members = peopleSearch.people.values;
    

    你必须给函数function displayPeopleSearch()添加一个参数

    function displayPeopleSearch(peopleSearch){
    
            var peopleSearchDiv = document.getElementById("peoplesearch");
    
            var members = peopleSearch.people.values;
    
    
            for (var member in members) {
            // but inside the loop, everything is the same
            // extract the title from the members first position
                peopleSearchDiv.innerHTML += "<p>" + members[member].firstName + " " + members[member].lastName + " is a " + members[member].positions.values[0].title + ".</p>";
    
            }
        }
    

    d.

    【讨论】:

    • 哦,愚蠢的错误,感谢@Didier指出!
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