Scala: How to get content from a REST URL (get REST content)

Here's a simple way to get content from a REST web service using Scala:

object GetUrlContent extends App {

  val url = "http://api.hostip.info/get_json.php?ip=12.215.42.19"
  val result = scala.io.Source.fromURL(url).mkString
  println(result)

}

That's a simple, "new" way I do it with Scala. However, note that it handles timeouts very poorly, such as if the web service you're calling is down or running slowly.

FWIW, here's an old approach I used to retrieve REST content (content from a REST URL):

  /**
   * Returns the text content from a REST URL. Returns a blank String if there
   * is a problem. (Probably should use Option/Some/None; I didn't know about it
   * back then.)
   */
  def getRestContent(url:String): String = {
    val httpClient = new DefaultHttpClient()
    val httpResponse = httpClient.execute(new HttpGet(url))
    val entity = httpResponse.getEntity()
    var content = ""
    if (entity != null) {
      val inputStream = entity.getContent()
      content = io.Source.fromInputStream(inputStream).getLines.mkString
      inputStream.close
    }
    httpClient.getConnectionManager().shutdown()
    return content
  }

This function just returns the content as a String, and then you can extract whatever you need from the String once you have it.

I don't have time to condense all my content right now, but if you search the website for other REST examples, you'll see how to properly handle timeouts when calling a REST web service.

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