我有一个电影供稿,我正在尝试检索具有特定键值的对象的“名称”值。这是回复:
{
genres: [
{
id: 18,
name: "Drama"
},
],
homepage: "http://www.therewillbeblood.com/",
id: 7345,
imdb_id: "tt0469494",
release_date: "2007-12-26",
crew: [
{
id: 4762,
name: "Paul Thomas Anderson",
department: "Writing",
job: "Screenplay",
profile_path: "/6lDE5N6lQUvAbKBRazn2Q0mRk44.jpg"
},
{
id: 52563,
name: "Upton Sinclair",
department: "Writing",
job: "Novel",
profile_path: null
},
{
id: 2950,
name: "Robert Elswit",
department: "Camera",
job: "Director of Photography",
profile_path: null
},
{
id: 1809,
name: "Dylan Tichenor",
department: "Editing",
job: "Editor",
profile_path: null
},
{
id: 4762,
name: "Paul Thomas Anderson",
department: "Directing",
job: "Director",
profile_path: "/6lDE5N6lQUvAbKBRazn2Q0mRk44.jpg"
},
{
id: 4772,
name: "Cassandra Kulukundis",
department: "Production",
job: "Casting",
profile_path: null
}
]
}
所以我试图用“导演”的“工作”键值获得该对象,这将是Paul Thomas Anderson。
这是我的javascript,我根据release_date的值设置年变量。试着找出如何设置导演变量:
<script type="text/javascript">
$(document).ready(function() {
var url = 'http://api.themoviedb.org/3/movie/';
imdb_id = 'tt0102685';
key = '?api_key=XXXXXXXXXXXX';
append = '&append_to_response=credits';
$.ajax({
type: 'GET',
url: url + imdb_id + key + append,
dataType: 'jsonp',
success: function(data) {
var $year = data.release_date.substring(0, 4);
$('#year').html($year);
},
});
});
</script>
任何帮助将不胜感激!花了将近两天的时间试图弄明白无济于事。
答案 0 :(得分:3)
如果您使用的是现代浏览器,则可以使用filter功能:
var name = data.crew.filter(function(e){
return (e.job=="Director") ? true:false;
})[0].name;
JS小提琴: http://jsfiddle.net/gTzyT/
错误处理示例
function getDirectorName(crew){
var name = "";
try{
name = json.crew.filter(function(e){
return e.job=="Director"
})[0].name;
}catch(err){
name = "No Director";
}
return name;
}
//calling the method
alert(getDirectorName(data.crew));
答案 1 :(得分:0)
对于稍微“不同”的方法,首先按作业排序。 E.g。
data.crew.sort(function(a,b) {
return a.job == 'Director' ? -1 : 1;
});
console.log(data.crew[0].name); // => name of Director
注意:我不一定赞同这个解决方案。排序不是最高效的方法。它有副作用(改变data.crew
数组的顺序,这可能是不可取的)。但这是一种“可爱”的解决方案,所以我想我会把它丢在那里。 :)