按字母顺序比较字符串

Question

String s1 = "Project";
String s2 = "Sunject";

我想按字母顺序比较上面两个字符串(在本例中为"Project",然后是"Sunject","P"在"S"之前).有谁知道如何用Java做到这一点？

Answer 1

String.compareTo 可能是也可能不是你需要的.

如果您需要字符串的本地化排序,请查看此链接.

Answer 2

看看这个String.compareTo方法.

s1.compareTo(s2)

来自javadocs:

如果此String对象按字典顺序位于参数字符串之前,则结果为负整数.如果此String对象按字典顺序跟随参数字符串,则结果为正整数.如果字符串相等,结果为零; compareTall在equals(Object)方法返回true时完全返回0.

Answer 3

String a = "..."; 
String b = "...";  

int compare = a.compareTo(b);  

if (compare < 0) {  
    //a is smaller
}
else if (compare > 0) {
    //a is larger 
}
else {  
    //a is equal to b
} 

Answer 4

您可以调用string的compareTo方法(java.lang.String.compareTo).java文档站点上详细记录了此功能.

这是一个简短的程序来演示它:

class StringCompareExample {
    public static void main(String args[]){
        String s1 = "Project"; String s2 = "Sunject";
        verboseCompare(s1, s2);
        verboseCompare(s2, s1);
        verboseCompare(s1, s1);
    }

    public static void verboseCompare(String s1, String s2){
        System.out.println("Comparing \"" + s1 + "\" to \"" + s2 + "\"...");

        int comparisonResult = s1.compareTo(s2);
        System.out.println("The result of the comparison was " + comparisonResult);

        System.out.print("This means that \"" + s1 + "\" ");
        if(comparisonResult < 0){
            System.out.println("lexicographically precedes \"" + s2 + "\".");
        }else if(comparisonResult > 0){
            System.out.println("lexicographically follows \"" + s2 + "\".");
        }else{
            System.out.println("equals \"" + s2 + "\".");
        }
        System.out.println();
    }
}

这是一个现场演示,显示它的工作原理:http://ideone.com/Drikp3

Answer 5

对于国有化后的字母顺序，请使用Collator。

//Get the Collator for US English and set its strength to PRIMARY
Collator usCollator = Collator.getInstance(Locale.US);
usCollator.setStrength(Collator.PRIMARY);
if( usCollator.compare("abc", "ABC") == 0 ) {
    System.out.println("Strings are equivalent");
}

有关支持的语言环境的列表，请参见JDK 8和JRE 8支持的语言环境。