How to compare strings alphabetically in Java?

@elwin_wilderman To compare two strings alphabetically in Java, you can use the compareTo() method. This method returns an integer value that indicates whether the first string comes before, after, or is the same as the second string in the dictionary. Here's an example:

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String str1 = "apple";
String str2 = "banana";

int result = str1.compareTo(str2);

if (result < 0) {
 // str1 comes before str2 alphabetically
} else if (result > 0) {
 // str1 comes after str2 alphabetically
} else {
 // str1 is the same as str2 alphabetically
}

Note that the compareTo() method is case-sensitive, so "Apple" would be considered to come after "banana" in the dictionary. To perform a case-insensitive comparison, you can convert both strings to lowercase (or uppercase) using the toLowerCase() method before calling compareTo().

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String str1 = "Apple";
String str2 = "banana";

// Convert both strings to lowercase before comparing
int result = str1.toLowerCase().compareTo(str2.toLowerCase());

@elwin_wilderman 

In Java, you can compare strings alphabetically using the compareTo() method, which is a method of the String class. The compareTo() method returns an integer value that indicates the relationship between two strings. Here's an example:

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String str1 = "apple";
String str2 = "banana";
int result = str1.compareTo(str2);

if (result < 0) {
    System.out.println("str1 comes before str2");
} else if (result > 0) {
    System.out.println("str2 comes before str1");
} else {
    System.out.println("str1 and str2 are equal");
}

In this example, we compare the strings "apple" and "banana" using the compareTo() method. The result of the comparison is stored in the result variable. If result is less than 0, it means that str1 comes before str2 alphabetically. If result is greater than 0, it means that str2 comes before str1 alphabetically. If result is equal to 0, it means that str1 and str2 are equal.