The compareTo() method compares two strings lexicographically.
The comparison is based on the Unicode value of each character in the strings.
You need to call length() method for each string to compare lengths
if(one.length() > two.length()) ...
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String::length is a method reference. Using it is equivalent to
Comparator.comparing(s -> s.length())
So it compares strings by comparing their length.
TL;DR: You are passing a method reference
Java 8 introduces lambda syntax to the Java language, this is a small part of that, it allows you to pass a method as a lambda.
So you call the method Comparator.comparing, this requires a "key extractor" function that takes an object from the Collection<T> and returns some object of a type U extends Comparable<? super U>.
The method String.length returns an Integer (when boxed) and an Integer is comparable to another Integer. So the call Comparator.comparing(String::length) returns a Comparator<String> that:
- takes a pair of
String,s1, s2 - calls
l1 = s1.lengthandl2 = s2.length - returns
l1.compareTo(l2)
So, Why there is no parenthesis?
Because you are passing a method reference. This method will be called later on an as yet undefined instance. Notice that length is not static on String so String.length() would not compile.
A simple solution is to iterate over all names and compare the length using the .length() method.
Another solution is to compare the length each time you enter a string (only one for loop). Here's an example :
int nb = 2;
String[] names = new String[nb];
int maxLength = -1; // Default value
for (int i = 0; i < nb; i++) {
names[i] = JOptionPane.showInputDialog("Enter your name : ");
// Compare the length of this name to the previous greatest length
if (names[i].length() > maxLength) {
// Change the maximum
maxLength = names[i].length();
}
}
JOptionPane.showMessageDialog(null, "The greatest value is " + maxLength);
You can iterate all the names and use the .length()method in String:
String longest = "";
for (String name : names) {
longest = name.length() > longest.length() ? name : longest;
}
This will iterate through your name list (names) and assign the longest to the variable longest
You need to use:
test.equals("example")
Because you are checking if the objects are equal, not if the object values are equal.
This is a pretty common mistake, but thankfully it has a very simple fix. :)
mainly in java we don't compare String with == operator since String is class we use equals method:
test.equals("");
it returns true if the two strings are equal and false if not
== tests for reference equality (whether they are the same object).
.equals() tests for value equality (whether they contain the same data).
Objects.equals() checks for null before calling .equals() so you don't have to (available as of JDK7, also available in Guava).
Consequently, if you want to test whether two strings have the same value you will probably want to use Objects.equals().
// These two have the same value
new String("test").equals("test") // --> true
// ... but they are not the same object
new String("test") == "test" // --> false
// ... neither are these
new String("test") == new String("test") // --> false
// ... but these are because literals are interned by
// the compiler and thus refer to the same object
"test" == "test" // --> true
// ... string literals are concatenated by the compiler
// and the results are interned.
"test" == "te" + "st" // --> true
// ... but you should really just call Objects.equals()
Objects.equals("test", new String("test")) // --> true
Objects.equals(null, "test") // --> false
Objects.equals(null, null) // --> true
From the Java Language Specification JLS 15.21.3. Reference Equality Operators == and !=:
While
==may be used to compare references of typeString, such an equality test determines whether or not the two operands refer to the sameStringobject. The result isfalseif the operands are distinctStringobjects, even if they contain the same sequence of characters (§3.10.5, §3.10.6). The contents of two stringssandtcan be tested for equality by the method invocations.equals(t).
You almost always want to use Objects.equals(). In the rare situation where you know you're dealing with interned strings, you can use ==.
From JLS 3.10.5. String Literals:
Moreover, a string literal always refers to the same instance of class
String. This is because string literals - or, more generally, strings that are the values of constant expressions (§15.28) - are "interned" so as to share unique instances, using the methodString.intern.
Similar examples can also be found in JLS 3.10.5-1.
Other Methods To Consider
String.equalsIgnoreCase() value equality that ignores case. Beware, however, that this method can have unexpected results in various locale-related cases, see this question.
String.contentEquals() compares the content of the String with the content of any CharSequence (available since Java 1.5). Saves you from having to turn your StringBuffer, etc into a String before doing the equality comparison, but leaves the null checking to you.
== tests object references, .equals() tests the string values.
Sometimes it looks as if == compares values, because Java does some behind-the-scenes stuff to make sure identical in-line strings are actually the same object.
For example:
String fooString1 = new String("foo");
String fooString2 = new String("foo");
// Evaluates to false
fooString1 == fooString2;
// Evaluates to true
fooString1.equals(fooString2);
// Evaluates to true, because Java uses the same object
"bar" == "bar";
But beware of nulls!
== handles null strings fine, but calling .equals() from a null string will cause an exception:
String nullString1 = null;
String nullString2 = null;
// Evaluates to true
System.out.print(nullString1 == nullString2);
// Throws a NullPointerException
System.out.print(nullString1.equals(nullString2));
So if you know that fooString1 may be null, tell the reader that by writing
System.out.print(fooString1 != null && fooString1.equals("bar"));
The following are shorter, but it’s less obvious that it checks for null:
System.out.print("bar".equals(fooString1)); // "bar" is never null
System.out.print(Objects.equals(fooString1, "bar")); // Java 7 required