当缺少必填字段时，是否可以使 json4s 不抛出异常? [英] Is it possible to make json4s not to throw exception when required field is missing?

问题描述

是否可以让json4s在缺少必填字段时不抛出异常?

Is it possible to make json4s not to throw exception when required field is missing ?

当我从原始 json 字符串中提取"对象时，它会抛出这样的异常

When I "extract" object from raw json string it throws exception like this one

org.json4s.package$MappingException: No usable value for pager
No usable value for rpp
Did not find value which can be converted into byte
    at org.json4s.reflect.package$.fail(package.scala:98)
    at org.json4s.Extraction$ClassInstanceBuilder.org$json4s$Extraction$ClassInstanceBuilder$$buildCtorArg(Extraction.scala:388)
    at org.json4s.Extraction$ClassInstanceBuilder$$anonfun$11.apply(Extraction.scala:396)

是否可以让它为空?

推荐答案

很简单，你要使用 Option 和它的潜力，Some 和 None.

It's quite simple, you have to use Option and its potentials, Some and None.

val json = ("name" -> "joe") ~ ("age" -> Some(35));
val json = ("name" -> "joe") ~ ("age" -> (None: Option[Int]))

请注意，在上述情况下，将为您的 Option 执行 match.如果是None，则将其从字符串中完全移除，因此不会反馈null.

Beware though, in the above case a match will be performed for your Option. If it's None, it will be completely removed from the string, so it won't feed back null.

在相同的模式中，要解析不完整的 JSON，您可以使用带有 Option 的 case 类.

In the same pattern, to parse incomplete JSON, you use a case class with Option.

case class someModel(
    age: Option[Int],
    name: Option[String]
);
val json = ("name" -> "joe") ~ ("age" -> None);
parse(json).extract[someModel];

有一种方法不会抛出任何异常，那就是extractOpt

There is a method which won't throw any exception, and that is extractOpt

parse(json).extractOpt[someModel];

使用 scala API 复制它的一种方法是使用 scala.util.Try:

A way to replicate that with the scala API would be to use scala.util.Try:

Try { parse(json).extract[someModel] }.toOption

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