当前，我能够检索Netflix提供的节目列表。
但是我的分数过滤器没有返回我想要的。
我想返回tmdb：score为8或更大的列表。
由于数据结构（存储在searchResult变量中）正在拆分provider_type和value，因此我现在得到的得分都在8分以上。
有没有办法做到这一点？ 如果可能的话，将其与我的Netflix提供程序功能结合以返回tmdb：score为8或更高的Netflix节目？
谢谢。
Node JS文件
const JustWatch = require("justwatch-api")
const netflixId = 8;
function print_result (name, result) {
console.log(name+":");
console.log(JSON.stringify(result, null, 4));
console.log("\n\n\n\n");
}
(async function(){
var justwatch = new JustWatch();
var searchResult = await justwatch.search('a');
searchResult.items.map(function(movie){
movie.offers.forEach(offer => {
if(offer.provider_id === netflixId){
print_result("search", searchResult)
}
})
})
searchResult.items.map(function(movie){
movie.scoring.filter(score =>{
if(score.provider_type ==="tmdb:score" && score.value > 8) {
print_result("search", searchResult)
}
})
})
}) ();
数据结构（存储在searchResult变量中，该对象包含一个计数total_results和一个对象数组items，每个对象代表一部电影）
{
"total_results": 27476,
"items": [
{
"jw_entity_id": "ts80908",
"id": 80908,
"title": "A Very English Scandal",
"full_path": "/us/tv-show/a-very-english-scandal",
"full_paths": {
"SHOW_DETAIL_OVERVIEW": "/us/tv-show/a-very-english-scandal"
},
"poster": "/poster/60368458/{profile}",
"original_release_year": 2018,
"tmdb_popularity": 3.422,
"object_type": "show",
"offers": [
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 9,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "sd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
},
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 9,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "hd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
}
],
"scoring": [
{
"provider_type": "imdb:score",
"value": 7.8
},
{
"provider_type": "tmdb:popularity",
"value": 3.422
},
{
"provider_type": "tmdb:score",
"value": 8.1
}
]
},
{
"jw_entity_id": "tm205151",
"id": 205151,
"title": "Alpha",
"full_path": "/us/movie/alpha-2015",
"full_paths": {
"MOVIE_DETAIL_OVERVIEW": "/us/movie/alpha-2015"
},
....
答案 0 :(得分：0)
据我了解，您想过滤tmdb：score为8或更高的Netflix节目吗？那么我们可以只使用filter来过滤项目吗？
searchResult.items.filter(function(movie){
var scorings = movie.scoring;
var filterScores =
scorings.filter(score => score.provider_type ==="tmdb:score" && score.value > 8);
return filterScores.length > 0;
})
答案 1 :(得分：0)
如果您只想过滤一个tmdb：score> = 8.0的电影对象数组，请使用以下代码段进行演示（请注意，我使用items数组来模拟与您相同格式的对象只是两个几乎完全相同的对象，一个用于演示目的，其得分> 8.0，另一个与得分<8.0，得分：
const searchResult = {
"total_results": 27476,
"items": [
{
"jw_entity_id": "ts80908",
"id": 80908,
"title": "A Very English Scandal",
"full_path": "/us/tv-show/a-very-english-scandal",
"full_paths": {
"SHOW_DETAIL_OVERVIEW": "/us/tv-show/a-very-english-scandal"
},
"poster": "/poster/60368458/{profile}",
"original_release_year": 2018,
"tmdb_popularity": 3.422,
"object_type": "show",
"offers": [
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 9,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "sd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
},
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 9,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "hd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
}
],
"scoring": [
{
"provider_type": "imdb:score",
"value": 7.8
},
{
"provider_type": "tmdb:popularity",
"value": 3.422
},
{
"provider_type": "tmdb:score",
"value": 8.1
}
]
},{
"jw_entity_id": "ts80909",
"id": 80908,
"title": "Some Other Title",
"full_path": "/us/tv-show/a-very-english-scandal",
"full_paths": {
"SHOW_DETAIL_OVERVIEW": "/us/tv-show/a-very-english-scandal"
},
"poster": "/poster/60368458/{profile}",
"original_release_year": 2018,
"tmdb_popularity": 3.422,
"object_type": "show",
"offers": [
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 9,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "sd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
},
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 9,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "hd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
}
],
"scoring": [
{
"provider_type": "imdb:score",
"value": 7.8
},
{
"provider_type": "tmdb:popularity",
"value": 3.422
},
{
"provider_type": "tmdb:score",
"value": 7.9
}
]
}]
}
const desiredMovies = searchResult.items.filter(movie => {
let tmdbScoring = movie.scoring.find(scoringObj => scoringObj.provider_type==="tmdb:score");
return tmdbScoring && tmdbScoring.value>=8.0;
})
console.log(desiredMovies);
此外，我注意到您添加的代码中似乎存在问题，这可能解释了为什么要获得所有电影。
在下面的代码中，您为发现的每个成功结果打印一次searchResult，这是一个包含计数和整个数组items的对象，因此将显示您可以看到数据结构中的所有电影。
//searchResult is the object that contains the entire array "items"
searchResult.items.map(function(movie){
movie.scoring.filter(score =>{
if(score.provider_type ==="tmdb:score" && score.value > 8) {
//the variable searchResult has scope here so it will contain the entire list
print_result("search", searchResult)
}
})
})
免责声明：以下内容仅是建设性的批评。
此外，我可以肯定您对.map和.filter的使用是不正确的。提供给这些函数的回调需要返回一个布尔值，以便.map或.filter函数本身可以返回一个新数组（要注意，.map和.filter不能影响原始数组）。提供的回调不仅不会返回布尔值，而且映射或过滤器本身的结果也不会存储在变量中。因此，我建议您看一下这些高阶函数的文档，以更好地了解它们的工作原理：
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/map
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/filter
因此，我建议您创建一个函数，该函数从对象的items列表中返回已过滤的电影列表，并一起打印。
答案 2 :(得分：0)
"SELECT
p.*,
pr.unlimited as 'unlimited'
FROM
`broadcast_portal` as p INNER JOIN broadcast_portal_registration as pr on p.id=pr.portal_id
WHERE
pr.group_id = 49
HAVING (unlimited = true or (select MAX(cpl._limit) FROM commercial_portal_limit as cpl INNER JOIN user on cpl.commercial_id=user.id INNER JOIN user_user_group uug ON user.id = uug.user_id WHERE uug.group_id=49 AND cpl.portal_id=p.id) > 0)
ORDER BY p.name ASC "
);
$qb->execute();
return $qb->fetchAll();
const netflixId = 8
const searchResult = {
"items": [
{
"offers": [
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 8,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "sd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
},
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 9,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "hd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
}
],
"scoring": [
{
"provider_type": "imdb:score",
"value": 7.8
},
{
"provider_type": "tmdb:popularity",
"value": 3.422
},
{
"provider_type": "tmdb:score",
"value": 8.1
}
]
},
{
"offers": [
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 8,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "sd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
},
{
"type": "aggregated",
"monetization_type": "flatrate",
"provider_id": 9,
"currency": "USD",
"subtitle_languages": [
"en"
],
"presentation_type": "hd",
"element_count": 1,
"new_element_count": 1,
"date_provider_id": "2019-09-14_9",
"date_created": "2019-09-14"
}
],
"scoring": [
{
"provider_type": "imdb:score",
"value": 7.8
},
{
"provider_type": "tmdb:popularity",
"value": 3.422
},
{
"provider_type": "tmdb:score",
"value": 7.9
}
]
}
]}
const movies = searchResult.items.filter(item => {
return item.offers.some(offer => offer.provider_id === netflixId) && item.scoring.some(score => score.provider_type === 'tmdb:score'&&score.value>=8)
})
console.log(movies)此过滤器功能可根据2个条件过滤出项目。
希望有帮助。