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Java Java Basics Perfecting the Prototype String Equality

I got the first if statement right, but I am struggling to understand why the second if statement does not work..

Struggling to understand why the second if statement does not work. It points to the "." after firstExample?

Equality.java
// I have imported a java.io.Console for you, it is named console. 
String firstExample = "hello";
String secondExample = "hello";
String thirdExample = "HELLO";

if (firstExample == secondExample)
{
  console.printf("first is equal to second");
}

if (firstExample.equalsIgnoreCase == thirdExample.equalsIgnoreCase)
{
  console.printf("first and third are the same ignoring case");
}
Simon Coates
Simon Coates
28,694 Points

equalsIgnoreCase is a method on string. You pass in the string for comparison as a parameter. (see below.)

The other thing wrong with this code is the use of ==. for more.

2 Answers

Simon Coates
Simon Coates
28,694 Points
// I have imported a java.io.Console for you, it is named console. 
String firstExample = "hello";
String secondExample = "hello";
String thirdExample = "HELLO";

if(firstExample.equals( secondExample)) {
  console.printf("first is equal to second");

}
if(firstExample.toLowerCase().equals( thirdExample.toLowerCase())) {
  console.printf("first and third are the same ignoring case");
}

java also has a equalsIgnoreCase method on string so you can bypass case conversion.

if(firstExample.equalsIgnoreCase( thirdExample)) {
  console.printf("first and third are the same ignoring case");
}

Because you're condition in the second if statement is wrong. equalsIgnoreCase() is a method, you will learn about methods in detail in the upcoming course. It is suppose to look like this

if (firstExample.equalsIgnoreCase( thirdExample)) { console.printf("first and third are the same ignoring case"); }

Hope it heped you understand .

Have great career!