Gracenote API调用以获取程序/节目

问题描述 投票:0回答:1

我试图通过Gracenote API搜索查询(show的名称)来获取程序。

供参考:http://developer.tmsapi.com/io-docs(v1.1 / programs / search)

使用他们网站上列出的示例(获取电影),它工作正常。

http://developer.tmsapi.com/Sample_Code

<html>
    <head>
        <style type="text/css">
            .tile {
                display: inline-block;
                border: 2px;
                padding: 4px;
                text-align: left;
                font-size: 15px;
                width:250px;
                font-family: Avenir;
                color: white;
            }
            </style>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
        <script>

            // construct the url with parameter values
            var apikey = "xxxxx";
            var baseUrl = "http://data.tmsapi.com/v1.1";
            var showtimesUrl = baseUrl + '/movies/showings';
            var zipCode = "78701";
            var d = new Date();
            var today = d.getFullYear() + '-' + (d.getMonth()+1) + '-' + d.getDate();

            $(document).ready(function() {

                              // send off the query
                              $.ajax({
                                     url: showtimesUrl,
                                     data: {    startDate: today,
                                     zip: zipCode,
                                     jsonp: "dataHandler",
                                     api_key: apikey
                                     },
                                     dataType: "jsonp",
                                     });
                              });

                              // callback to handle the results
                              function dataHandler(data) {
                                  $(document.body).append('<h2>Found ' + data.length + ' movies showing within 5 miles of ' + zipCode+'</h2>');
                                  $.each(data, function(index, movie) {
                                         var movieData = '<div class="tile"><img src="http://fanc.tmsimg.com/' + movie.preferredImage.uri + '?api_key='+apikey+'"><br/>';
                                         movieData += 'Title:' + movie.title + '<br>';
                                         movieData += 'ID: ' + movie.tmsId + '<br>';
                                         if (movie.ratings) {movieData += 'Rating: ' + movie.ratings[0].code;}
                                         else {movieData += 'Rating: ' + 'N/A';}
                                         $(document.body).append(movieData);
                                         });
                              }

            </script>
    </head>
    <body>
    </body>
</html>

当我试图修改它(以获取程序)时,我无法检索任何数据,所有数据都返回为未定义。

<html>
    <head>
        <style type="text/css">
            .tile {
                display: inline-block;
                border: 2px;
                padding: 4px;
                text-align: left;
                font-size: 15px;
                width:250px;
                font-family: Avenir;
                color: white;
            }
            </style>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
        <script>

            // construct the url with parameter values
            var apikey = "xxxxx";
            var baseUrl = "http://data.tmsapi.com/v1.1";
            var showtimesUrl = baseUrl + '/programs/search';
            var zipCode = "78701";
            var showName = 'Friends';

            $(document).ready(function() {

                              // send off the query
                              $.ajax({
                                     url: showtimesUrl,
                                     data: {    q: showName,
                                     jsonp: "dataHandler",
                                     api_key: apikey
                                     },
                                     dataType: "jsonp",
                                     });
                              });

                              // callback to handle the results
                              function dataHandler(data) {
                                  $(document.body).append('<h2>Found ' + data.length + ' movies showing within 5 miles of ' + zipCode+'</h2>');
                                  var programs = data.hits;
                                  $.each(data, function(index, program) {
                                         var programData = '<div class="tile">' + program.entityType + '<br/>';
                                         programData += 'Title:' + program.title + '<br>';
                                         programData += 'ID: ' + program.tmsId + '<br>';
                                         $(document.body).append(programData);
                                         });
                              }

            </script>
    </head>
    <body>
    </body>
</html>

实际结果:(响应状态:200)

未定义

标题:未定义

ID:未定义

预期结果:(响应状态:200)

节目

标题:朋友们

ID:SH001151270000

javascript jquery json jsonp
1个回答
0
投票

我能够通过查看数组来解决它。

对于网站提供的电影示例代码,数组返回属性路径为[“”0“”]。entityType但是对于Program,它返回为命中[“”0“”]。program.entityType。

所以我用data.hits替换了数据,用program.program.entityType等替换了program.entityType。

  function dataHandler(data) {
    $(document.body).append('<h2>Found ' + data.hits.length + ' shows.</h2>');
    $.each(data.hits, function(index, program) {
      var programData = '<div class="tile">' + program.program.entityType + '<br/>';
      programData += 'Title:' + program.program.title + '<br>';
      programData += 'ID: ' + program.program.tmsId + '<br>';
      $(document.body).append(programData);
    });
  }

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