NOTE: This is an old article about how to make an HTTP GET request using Scala. Please note that there are plenty of HTTP/HTTPS libraries in the world now, so I’m only keeping this article here for historical purposes.
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.
