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什么最直截了当的办法是在Schala教的JSON?
原标题:What is the most straightforward way to parse JSON in Scala?

I m与Schala一道,在网上应用。 该计划是从外部APIC获得JSON数据,并插入一个模板(很幸运的是,在XML中获取数据并非一种选择)。

我先尝试与Twitter scala-json图书馆合作,但我无法适当汇编(《关于兴奋剂的守则》没有在纸浆中更新,说没有标准项目7.10,我没有做过。

升华会看起来令人印象深刻,但似乎比我现在更详尽。

试图进口一个图书馆一号楼,在Java, j,造成各种电弧错误。 这太糟糕,因为我喜欢直截了当的onic。

我在Scala.util.parsing.json建造工程取得了一定的进展。 JSON,但实际上我可以告诉如何接触这些要素。 我对Schala说几句新话,正如你可能指出的那样。 你们如何获得初等目标的财产?

scala.util.parsing.json.JSON 有许多信息,但对于如何在任何地方利用这些信息,是否有直截了当的理论?

我真的只想把眼光降为仁、歌、地图和名单。 我无需将物体序列化,或使被 des化物体在当前被归入一个类别。

谁能向我指出如何与上述图书馆之一合作,或者帮助我用 Java语建立,我将做些什么?

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