2012-11-01 69 views
1

在我们的Spotify APP中,我们正在调用model.Search函数,以便我们验证最多100位艺术家的列表。问题是我们可能会验证的艺术家之一可能是“!!!”。这是一个真实的艺术家姓名,但它会导致Spotify返回,验证艺术家姓名的错误代码和空列表。我们可以在应用程序中一一验证艺术家姓名,因此可以处理错误代码,但这会大大降低应用程序的速度。这是API中的错误吗? Spotify是否会尽快解决它?以下是我们正在拨打的电话的一个示例:Spotify API在模型上返回错误。搜索艺术家“!!!”

var search = new models.Search("artist:"Mother Love Bone" OR artist:"Hole" OR artist:"Soundgarden" OR  artist:"Stone Temple Pilots" OR artist:"Blind Melon" OR artist:"Silverchair" OR artist:"Alice In Chains" OR  artist:"Bush" OR artist:"Nirvana" OR artist:"Pearl Jam" OR artist:"Big Black" OR artist:"Hüsker Dü" OR  artist:"Throwing Muses" OR artist:"Frank Black" OR artist:"Fugazi" OR artist:"Santogold" OR artist:"This  Mortal Coil" OR artist:"Pixies" OR artist:"Lykke Li" OR artist:"Dinosaur Jr." OR artist:"Dinosaur Jr" OR  artist:"Morphine" OR artist:"Guided By Voices" OR artist:"Spacemen 3" OR artist:"Ride" OR artist:"My  Bloody Valentine" OR artist:"Destroyer" OR artist:"Slowdive" OR artist:"Spiritualized" OR artist:"Wavves"  OR artist:"Joanna Newsom" OR artist:"The Beta Band" OR artist:"The Jon Spencer Blues Explosion" OR  artist:"Jon Spencer Blues Explosion" OR artist:"Death From Above 1979" OR artist:"Buzzcocks" OR  artist:"The Buzzcocks" OR artist:"Television" OR artist:"Battles" OR artist:"Morrissey" OR artist:"The Stone  Roses" OR artist:"Ian Brown" OR artist:"The Libertines" OR artist:"The Psychedelic Furs" OR  artist:"Psychedelic Furs" OR artist:"Kings Of Convenience" OR artist:"Happy Mondays" OR artist:"Yo La  Tengo" OR artist:"Echo & The Bunnymen" OR artist:"The Specials" OR artist:"Interpol" OR artist:"Yeah  Yeah Yeahs" OR artist:"Doves" OR artist:"Dover" OR artist:"Joy Division" OR artist:"Silver Jews" OR  artist:"!!!""); 
search.localResults = models.LOCALSEARCHRESULTS.IGNORE; 
search.searchArtists = true; 
search.searchAlbums = false; 
search.searchTracks = false; 
search.searchPlaylists = false; 
search.observe(models.EVENT.CHANGE, function() { ... }); 
search.observe(models.EVENT.LOAD_ERROR, function() { // We got error here }); 
search.appendNext(); 
+0

错误,试图使该代码块,它失败 – Woot4Moo

回答

0

我无法在Web API搜索(我有权访问的所有内容)中复制该内容。 “!!!”不会返回任何结果,但也不会返回错误。

也许尝试URL编码它?那会变成“!!!”变成“%21%21%21”。它在Web API中没有任何区别 - 仍然没有返回结果,但没有错误。但是可能存在特定于主API的编码故障。

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