String Comparison in Java

How do I compare strings in Java?

== 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

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 method String.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.

Wrong results while String comparison

To compare two strings u should use the method equals() or equalsIgnoreCase().

in your case:

if(it.next().equals(args[0]))

the operator == returns true if the two object are the same object, same address in memory.

Java String comparison best practice

I prefer the Yoda Expression "name".equals(person.getName()); since it means you don't need to check if person.getName() is null. That saves a bit of typing and is arguably clearer once you get used to it.

Although in your case, you'll still need to check if person is not null.

String Comparison in Java

Leading from answers from @Bozho and @aioobe, lexicographic comparisons are similar to the ordering that one might find in a dictionary.

The Java String class provides the .compareTo () method in order to lexicographically compare Strings. It is used like this "apple".compareTo ("banana").

The return of this method is an int which can be interpreted as follows:

• returns < 0 then the String calling the method is lexicographically first (comes first in a dictionary)
• returns == 0 then the two strings are lexicographically equivalent
• returns > 0 then the parameter passed to the compareTo method is lexicographically first.

More specifically, the method provides the first non-zero difference in ASCII values.

Thus "computer".compareTo ("comparison") will return a value of (int) 'u' - (int) 'a' (20). Since this is a positive result, the parameter ("comparison") is lexicographically first.

There is also a variant .compareToIgnoreCase () which will return 0 for "a".compareToIgnoreCase ("A"); for example.

String Comparison not working in Java

Try this , as this is the recommended way of doing it

if (!(rs.getString("order_status")).equals(ordstat))

Note that == compares references of the objects and if you want to compare actual content of strings , use equals() function

String comparison in Java

You use 'equals' when you compare strings.

if(s1.equals(s2)) { true }

Difference in the results of String comparison in Java?

"Chai"+"pau" is semantically identical to "Chaipau", and thus is the same instance that a refers to.

"Chai"+b is evaluated at runtime (because b is not a compile-time constant expression), creating a new instance of String, and thus is not the same instance that a refers to.

String.equals versus ==

Use the string.equals(Object other) function to compare strings, not the == operator.

The function checks the actual contents of the string, the == operator checks whether the references to the objects are equal. Note that string constants are usually "interned" such that two constants with the same value can actually be compared with ==, but it's better not to rely on that.

if (usuario.equals(datos[0])) {
    ...
}

NB: the compare is done on 'usuario' because that's guaranteed non-null in your code, although you should still check that you've actually got some tokens in the datos array otherwise you'll get an array-out-of-bounds exception.

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