Techie Shah

Introduction In this post, we will explore how to convert an array of bytes i.e. byte[] to String in Java. For this functionality, we will use the following constructor of java.lang.String class String (byte[] bytes,  String  enc) Example import java.nio.charset.StandardCharsets; public class ByteArrayToString { public static void main(String[] args) { byte[] bytes = "a quick fox jumps over the lazy dog".getBytes(); // convert the bytes to String System.out.println("bytes = " + bytes); System.out.println( "bytes.toString() = " + bytes.toString()); String myString = new String ( bytes , StandardCharsets.UTF_8); System.out.println("myString = " + myString); } } Output bytes = [B@41a4555e bytes.toString() = [B@41a4555e myString = a quick fox jumps over the lazy dog Conclusion In this post, we learned how to convert an array of bytes to String.

Introduction In this post, we will learn how to convert String to InputStream in Java. Example import java.io.ByteArrayInputStream; import java.io.InputStream; public class StringToInputStream { public static void main(String[] args) { String text = "A quick fox jumps over a lazy dog"; try ( InputStream inputStream = convertToInputStream(text) ) { //TODO you business logic goes here } catch (Exception e) { e.printStackTrace(); } } private static InputStream convertToInputStream(String text) throws Exception { return new ByteArrayInputStream(text.getBytes()); } } Conclusion This is the simplest and shortest way to convert a String to InputStream in Java.

Introduction Mostly we make a connection to Oracle using JDBC through SID. Recently I have to make a connection to Oracle using Service Name. I am writing this post to share my experience. SID Connection String  jdbc:oracle:thin:@IP:1521:SID Service Name connection String  jdbc:oracle:thin:@(description=(address=(host=IP)(protocol=tcp)(port=1521))(connect_data=(service_name=UR_SERVICE_NAME))) Replace Service Name and host address per your configuration These strings work both for plain JDBC and hibernate. For hibernate configuration, put the above string inside <property name="connection.url">Above connection String</property> tag of hibernate.cfg.xml. Conclusion In this post, we learn how to connect to Oracle using both SID and Service Name. Happy learning

Introduction In this tutorial, we will learn how to convert InputStream to String in Java. Example import java.io.InputStream; import java.net.URL; public class InputStreamToString { public static void main(String[] args) { //very frequent asked question in interview , regarding closable interface try(InputStream inputStream = new URL("https://www.oracle.com/index.html").openStream()){ System.out.println(convertToString(inputStream)); }catch (Exception e) { // TODO: handle exception } } private static String convertToString(InputStream inputStream) throws Exception{ return new String(inputStream.readAllBytes()); } } Conclusion Above example is the simplest method to convert an Inpustream to String in Java. There are alternate solutions available like using commonio jar and converting InputStream to ByteStream etc. Hope you enjoy this tutorial. Happy learning 

Introduction In this tutorial, we will learn how to skip or display % in a String using String.format MyStringFormatter .java  public class MyStringFormatter { public static void main(String[] args) { System.out.println(String.format( "Your score is %d % " ,65) ); } } Output Exception in thread "main" java.util.UnknownFormatConversionException: Conversion = ' ' at java.base/java.util.Formatter.checkText(Formatter.java:2732) at java.base/java.util.Formatter.parse(Formatter.java:2718) at java.base/java.util.Formatter.format(Formatter.java:2655) at java.base/java.util.Formatter.format(Formatter.java:2609) at java.base/java.lang.String.format(String.java:2897) at MyStringFormatter.main(MyStringFormatter.java:4) Solution To display a % in String when using String.format , you have to escape a % with % i.e. %% public class MyStringFormatter { public static void main(String[] args) { System.out.println(String.fo...

Exception  Exception in thread "main" java.lang.BootstrapMethodError: java.lang.NoClassDefFoundError: scala/runtime/java8/JFunction1$mcII$sp         at SparkPi$.main(SparkPi.scala:14)         at SparkPi.main(SparkPi.scala)         at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)         at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:62) Problem: It seems that you have compiled and generated your jar file with an upper version of the Scala compiler than the one spark is using. Solution Step 1 Run the following command from spark-installed-directory\bin spark-shell.cmd (or.sh) and note the Scala version, My Spark version was 2.4.3 and Scala version 2.11.12 Step 2 Change the scala version into your build.sbt to 2.11.12 (per your configuration). My build.sbt is name := "SparkPi Project" version := "1.0" scalaVersion := "...

Introduction  Currently, Apache Spark does not support JAVA 11 and later versions. When you try to run the Spark application, you may get the following exception Exception pyspark.sql.utils.IllegalArgumentException: 'Unsupported class file major version 55' Solution There are different ways to fix this exception like Set environmental variable for JAVA_HOME Modify Apache spark environment configuration file i.e. spark-env.sh or spark-env.cmd In this post, I will help you to set JAVA_HOME using Spark's configuration file Windows Environment  Go to the spark-directory\ conf Create a file by the name of spark-env.cmd  Paste the following line spark-env.cmd                   set JAVA_HOME=C:\Program Files\Java\jdk1.8.0_201 Linux and Mac Go to the spark-directory\ conf Open spark-env.sh Paste the following line spark-env.cmd        export JAVA_HOME=$(user/J...