Fix code that assumed spark.local.dir is only a single directory

core/src/main/scala/spark/Utils.scala

@@ -134,7 +134,7 @@ private object Utils extends Logging {
    */
   def fetchFile(url: String, targetDir: File) {
     val filename = url.split("/").last
-    val tempDir = System.getProperty("spark.local.dir", System.getProperty("java.io.tmpdir"))
+    val tempDir = getLocalDir
     val tempFile =  File.createTempFile("fetchFileTemp", null, new File(tempDir))
     val targetFile = new File(targetDir, filename)
     val uri = new URI(url)
@@ -204,6 +204,15 @@ private object Utils extends Logging {
     FileUtil.chmod(filename, "a+x")
   }
 
+  /**
+   * Get a temporary directory using Spark's spark.local.dir property, if set. This will always
+   * return a single directory, even though the spark.local.dir property might be a list of
+   * multiple paths.
+   */
+  def getLocalDir: String = {
+    System.getProperty("spark.local.dir", System.getProperty("java.io.tmpdir")).split(',')(0)
+  }
+
   /**
    * Shuffle the elements of a collection into a random order, returning the
    * result in a new collection. Unlike scala.util.Random.shuffle, this method

core/src/main/scala/spark/broadcast/HttpBroadcast.scala

@@ -89,7 +89,7 @@ private object HttpBroadcast extends Logging {
   }
 
   private def createServer() {
-    broadcastDir = Utils.createTempDir(System.getProperty("spark.local.dir", System.getProperty("java.io.tmpdir")))
+    broadcastDir = Utils.createTempDir(Utils.getLocalDir)
     server = new HttpServer(broadcastDir)
     server.start()
     serverUri = server.uri