我有一个简单的JS表单,它接受一个搜索词和一个下拉选项,并根据这些参数返回JSON数据。我相信我遇到的问题是即使intelliJ看到对另一个文件的引用,它也无法处理它('意外令牌错误')。JS从本地文件读取JSON对象
jsFiddle JSON片段:本
var songs = [
{"title":"12-Bar Original" , "writer":"Lennon, McCartney, Harrison and Starkey ", "vocalist":"Instrumental "},
{"title":"Across the Universe" , "writer":"Lennon ", "vocalist":"Lennon "},
{"title":"Act Naturally" , "writer":"Russell, Morrison ", "vocalist":"Starkey "},
{"title":"Ain't She Sweet" , "writer":"Yellen, Ager ", "vocalist":"Lennon "},
{"title":"All I've Got to Do" , "writer":"Lennon ", "vocalist":"Lennon "},
{"title":"All My Loving" , "writer":"McCartney ", "vocalist":"McCartney "},
{"title":"All Things Must Pass" , "writer":"Harrison — — ", "vocalist":" "},
{"title":"All Together Now" , "writer":"McCartney, with Lennon ", "vocalist":"McCartney, with Lennon "},
{"title":"All You Need Is Love" , "writer":"Lennon ", "vocalist":"Lennon "},
{"title":"And I Love Her" , "writer":"McCartney, with Lennon ", "vocalist":"McCartney "}];
的思考?我确信我错过了一些愚蠢的东西。欣赏眼睛。谢谢!
这是一个JS代码,而不是JSON。你如何使用它?通过ajax或smth加载?如果是这样,请删除“var songs =”并尾随“;”。 如果您想要运行JS代码 - 例如添加一个全局变量“歌曲”,你可以评估文件的文字。您还可以将代码修改为模块并通过RequireJS进行需求,或者只需将该文件加载到脚本标记中即可。 –
@PavelStaseljun - 不,它是完全有效的JSON。 – davidkonrad
Rly? JSON中的字符串“var songs =”代表什么? –