Hi Marcelo, Thank you for the prompt response. I tried adding listeners as well, didn't work either. Looks like it isn't starting the job at all.
[image: --] Tariq, Mohammad [image: https://]about.me/mti <https://about.me/mti?promo=email_sig&utm_source=email_sig&utm_medium=external_link&utm_campaign=chrome_ext> [image: http://] Tariq, Mohammad about.me/mti [image: http://] <http://about.me/mti> On Tue, Nov 8, 2016 at 5:06 AM, Marcelo Vanzin <van...@cloudera.com> wrote: > On Mon, Nov 7, 2016 at 3:29 PM, Mohammad Tariq <donta...@gmail.com> wrote: > > I have been trying to use SparkLauncher.startApplication() to launch a > Spark app from within java code, but unable to do so. However, same piece > of code is working if I use SparkLauncher.launch(). > > > > Here are the corresponding code snippets : > > > > SparkAppHandle handle = new SparkLauncher() > > > > .setSparkHome("/Users/miqbal1/DISTRIBUTED_WORLD/ > UNPACKED/spark-1.6.1-bin-hadoop2.6") > > > > .setJavaHome("/Library/Java/JavaVirtualMachines/jdk1.8.0_92 > .jdk/Contents/Home") > > > > .setAppResource("/Users/miqbal1/wc.jar").setMainClass("org. > myorg.WC").setMaster("local") > > > > .setConf("spark.dynamicAllocation.enabled", > "true").startApplication(); System.out.println(handle.getAppId()); > > > > System.out.println(handle.getState()); > > > > This prints null and UNKNOWN as output. > > The information you're printing is not available immediately after you > call "startApplication()". The Spark app is still starting, so it may > take some time for the app ID and other info to be reported back. The > "startApplication()" method allows you to provide listeners you can > use to know when that information is available. > > -- > Marcelo >
