如何使用 Play with Scala 加载 JSON 文件
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How to load JSON file using Play with Scala
提问by Henrique Ferrolho
I need to load a JSONfile with a list of cities in one of my controllers, in order to pass it to a viewafterwards.
我需要JSON在我的一个控制器中加载一个包含城市列表的文件，以便之后将其传递给视图。
I have placed the file here: app/assets/jsons/countriesToCities.json
(By the way: is this an appropriate location, or should I place it somewhere else?)
我把文件放在这里：（app/assets/jsons/countriesToCities.json
顺便说一句：这是一个合适的位置，还是我应该把它放在其他地方？）
I have read the docsand I can see it is possible to create a JsValuefrom a string: https://www.playframework.com/documentation/2.4.x/ScalaJson#Using-string-parsing
我已经阅读了文档，我可以看到可以JsValue从字符串创建一个：https: //www.playframework.com/documentation/2.4.x/ScalaJson#Using-string-parsing
I want to create a JsValuein a similar fashion. The difference is that I want to load the content from a file, not from a string... I haven't found any code snippet on how to do this, unfortunately.
Do I have to use something else to read the file into a string and only then use the parse method on that string?
我想以JsValue类似的方式创建一个。不同之处在于我想从文件中加载内容，而不是从字符串中加载...不幸的是，我还没有找到任何关于如何执行此操作的代码片段。
我是否必须使用其他方法将文件读入字符串，然后才对该字符串使用 parse 方法？
Code snippets with examples on how to do this in the answers will be highly appreciated! :)
非常感谢在答案中提供有关如何执行此操作的示例的代码片段！:)
Thank you very much in advance!
非常感谢您提前！
采纳答案by Henrique Ferrolho
Here is how I managed to solve it:
这是我设法解决它的方法：
val source: String = Source.fromFile("app/assets/jsons/countriesToCities.json").getLines.mkString
val json: JsValue = Json.parse(source)
Thanks for the help! :)
谢谢您的帮助！:)
回答by Dylan
Looks like the comment about the possible duplicate is how to read a file from your app/assets folder. My answer is about how to parse Json from a stream. Combine the two and you should be good to go.
看起来关于可能重复的评论是如何从您的 app/assets 文件夹中读取文件。我的答案是关于如何从流中解析 Json。将两者结合起来，你应该很高兴。
Json.parseaccepts a few different argument types, one of which is InputStream.
Json.parse接受几种不同的参数类型，其中之一是InputStream.
val stream = new FileInputStream(file)
val json = try { Json.parse(stream) } finally { stream.close() }
P.S. When you can't find what you're looking for in the written docs, the API Docsare a good place to start.
PS 当您在书面文档中找不到您要查找的内容时，API 文档是一个不错的起点。
回答by flurdy
From Play 2.6, Environment has the getExistingFile, getFile, resourceand resourceAsStreammethods, E.g.:
从播放2.6，环境有getExistingFile，getFile，resource和resourceAsStream方法，例如：
class Something @Inject (environment: play.api.Environment) {
// ...
environment.resourceAsStream("data.json") map ( Json.parse(_) )
(Note, in this case data.jsonis inside the conffolder)
（注意，在这种情况下data.json在conf文件夹中）
https://www.playframework.com/documentation/2.6.x/api/scala/index.html#play.api.Environment
https://www.playframework.com/documentation/2.6.x/api/scala/index.html#play.api.Environment
回答by mkarmona
I usually need few small dictionaries to live in memory in order to build some plain LUTs (LookUp Tables). In order to achieve that I use the following piece of code (please it is worth noting I am using latest Play 2.6):
我通常需要很少的小词典才能保存在内存中，以便构建一些普通的 LUT（查找表）。为了实现这一点，我使用了以下代码（请注意，我使用的是最新的Play 2.6）：
def loadJSONFromFilename(filename: String): Option[JsValue] =
Option(Source.fromFile(filename).mkString)
.map(Json.parse)
Using the previous function is just a matter of passing a filenamepath. Tailoring this to be used with Play you might need to place your file in a resource folder and to enable it you need to place this in your build.sbtfile:
使用前面的函数只是传递filename路径的问题。调整它以与 Play 一起使用，您可能需要将您的文件放在资源文件夹中，并启用它，您需要将它放在您的build.sbt文件中：
// include resources into the unversal zipped package using sbt dist
mappings in Universal ++= directory(baseDirectory.value / "resources")
Thus, you can use this resource folder this way in your Play class:
因此，您可以在 Play 类中以这种方式使用此资源文件夹：
lazy val lutFilePath: Path =
Paths.get(env.rootPath.getAbsolutePath, "resources", "lut.json")
lazy val lutJson = loadJSONFromFilename(lutFilePath.toString)
