How can i pass a URL explicitly in Scala - scala

Question

Hello i am new to Scala . I tried this code
def web ( url : Any) {
| val ur= new URL("url")
| val content=fromInputStream(ur.openStream).getLines.mkString("\n")
| print(content)
| }
when i pass a url like web("http://contentexplore.com/iphone-6-amazing-looks/")
it is showing error
java.net.MalformedURLException: no protocol: url
at java.net.URL.<init>(URL.java:585)
at java.net.URL.<init>(URL.java:482)
at java.net.URL.<init>(URL.java:431)
at .web(<console>:22)
at .<init>(<console>:23)
at .<clinit>(<console>)
at .<init>(<console>:11)
at .<clinit>(<console>)
at $print(<console>)
at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
at java.lang.reflect.Method.invoke(Method.java:601)
at scala.tools.nsc.interpreter.IMain$ReadEvalPrint.call(IMain.scala:704)
at scala.tools.nsc.interpreter.IMain$Request$$anonfun$14.apply(IMain.scala:920)
at scala.tools.nsc.interpreter.Line$$anonfun$1.apply$mcV$sp(Line.scala:43)
at scala.tools.nsc.io.package$$anon$2.run(package.scala:25)
at java.lang.Thread.run(Thread.java:722)
My question is how can i pass a url explicitly in scala .Kindly suggest me an idea .Thanks in advance

Answer 1

As mentioned in the comments, this line is the problem:
val ur= new URL("url")
If you want to create a URL from the input param url, the code should be:
val ur= new URL(url)
With the error, the java URL class was trying to parse a String with value "url", looking first for a recognized protocol (http, https, etc...) and not finding one, so that's why you were seeing that error.