也许有人会对我的解决方案感兴趣:

class Version private constructor(private val versionString: String) : Comparable<Version> {

    private val major: Int by lazy { versionString.split(".")[0].toInt() }

    private val minor: Int by lazy { versionString.split(".")[1].toInt() }

    private val patch: Int by lazy {
        val splitArray = versionString.split(".")

        if (splitArray.size == 3)
            splitArray[2].toInt()
        else
            0
    }

    override fun compareTo(other: Version): Int {
        return when {
            major > other.major -> 1
            major < other.major -> -1
            minor > other.minor -> 1
            minor < other.minor -> -1
            patch > other.patch -> 1
            patch < other.patch -> -1
            else -> 0
        }
    }

    override fun equals(other: Any?): Boolean {
        if (other == null || other !is Version) return false
        return compareTo(other) == 0
    }

    override fun hashCode(): Int {
        return major * minor * patch
    }

    companion object {
        private fun doesContainsVersion(string: String): Boolean {
            val versionArray = string.split(".")

            return versionArray.size in 2..3
                    && versionArray[0].toIntOrNull() != null
                    && versionArray[1].toIntOrNull() != null
                    && (versionArray.size == 2 || versionArray[2].toIntOrNull() != null)
        }

        fun from(string: String): Version? {
            return if (doesContainsVersion(string)) {
                Version(string)
            } else {
                null
            }
        }
    }
}

用法:

val version1 = Version.from("3.2")
val version2 = Version.from("3.2.1")
version1 <= version2

下面是一个优化的实现:

public static final Comparator<CharSequence> VERSION_ORDER = new Comparator<CharSequence>() {

  @Override
  public int compare (CharSequence lhs, CharSequence rhs) {
    int ll = lhs.length(), rl = rhs.length(), lv = 0, rv = 0, li = 0, ri = 0;
    char c;
    do {
      lv = rv = 0;
      while (--ll >= 0) {
        c = lhs.charAt(li++);
        if (c < '0' || c > '9')
          break;
        lv = lv*10 + c - '0';
      }
      while (--rl >= 0) {
        c = rhs.charAt(ri++);
        if (c < '0' || c > '9')
          break;
        rv = rv*10 + c - '0';
      }
    } while (lv == rv && (ll >= 0 || rl >= 0));
    return lv - rv;
  }

};

结果:

"0.1" - "1.0" = -1
"1.0" - "1.0" = 0
"1.0" - "1.0.0" = 0
"10" - "1.0" = 9
"3.7.6" - "3.7.11" = -5
"foobar" - "1.0" = -1

public int CompareVersions(String version1, String version2)
{
    String[] string1Vals = version1.split("\\.");
    String[] string2Vals = version2.split("\\.");

    int length = Math.max(string1Vals.length, string2Vals.length);

    for (int i = 0; i < length; i++)
    {
        Integer v1 = (i < string1Vals.length)?Integer.parseInt(string1Vals[i]):0;
        Integer v2 = (i < string2Vals.length)?Integer.parseInt(string2Vals[i]):0;

        //Making sure Version1 bigger than version2
        if (v1 > v2)
        {
            return 1;
        }
        //Making sure Version1 smaller than version2
        else if(v1 < v2)
        {
            return -1;
        }
    }

    //Both are equal
    return 0;
}

您需要规范化版本字符串，以便对它们进行比较。类似的

import java.util.regex.Pattern;

public class Main {
    public static void main(String... args) {
        compare("1.0", "1.1");
        compare("1.0.1", "1.1");
        compare("1.9", "1.10");
        compare("1.a", "1.9");
    }

    private static void compare(String v1, String v2) {
        String s1 = normalisedVersion(v1);
        String s2 = normalisedVersion(v2);
        int cmp = s1.compareTo(s2);
        String cmpStr = cmp < 0 ? "<" : cmp > 0 ? ">" : "==";
        System.out.printf("'%s' %s '%s'%n", v1, cmpStr, v2);
    }

    public static String normalisedVersion(String version) {
        return normalisedVersion(version, ".", 4);
    }

    public static String normalisedVersion(String version, String sep, int maxWidth) {
        String[] split = Pattern.compile(sep, Pattern.LITERAL).split(version);
        StringBuilder sb = new StringBuilder();
        for (String s : split) {
            sb.append(String.format("%" + maxWidth + 's', s));
        }
        return sb.toString();
    }
}

打印

'1.0' < '1.1' '1.0.1' < '1.1' '1.9' < '1.10' “1。A ' > '1.9'

最好的方法是重用现有代码， 使用Maven的ComparableVersion类

优点:

Apache许可证，版本2.0， 测试, 在多个项目中使用(复制)，如spring-security-core, jboss等 多个特性 它已经是java.lang。可比的了 只是复制粘贴一个类，没有第三方依赖

不要包含对maven-artifact的依赖项，因为那会拉动各种传递依赖项

我写了一个名为MgntUtils的开源库，它有一个用于字符串版本的实用程序。它正确地比较它们，适用于版本范围等等。下面是这个库javadoc参见方法TextUtils.comapreVersions(…)它已经被大量使用并经过了良好的测试。下面这篇文章描述了这个库以及如何获取它。它可以作为Maven工件和在github上获得(包括源代码和JavaDoc)