是否有比较版本号的标准习语?我不能直接使用String compareTo,因为我还不知道点释放的最大数量是多少。我需要比较版本,并有以下保持正确:

1.0 < 1.1
1.0.1 < 1.1
1.9 < 1.10

当前回答

这篇旧文章的另一个解决方案(对那些可能有帮助的人来说):

public class Version implements Comparable<Version> {

    private String version;

    public final String get() {
        return this.version;
    }

    public Version(String version) {
        if(version == null)
            throw new IllegalArgumentException("Version can not be null");
        if(!version.matches("[0-9]+(\\.[0-9]+)*"))
            throw new IllegalArgumentException("Invalid version format");
        this.version = version;
    }

    @Override public int compareTo(Version that) {
        if(that == null)
            return 1;
        String[] thisParts = this.get().split("\\.");
        String[] thatParts = that.get().split("\\.");
        int length = Math.max(thisParts.length, thatParts.length);
        for(int i = 0; i < length; i++) {
            int thisPart = i < thisParts.length ?
                Integer.parseInt(thisParts[i]) : 0;
            int thatPart = i < thatParts.length ?
                Integer.parseInt(thatParts[i]) : 0;
            if(thisPart < thatPart)
                return -1;
            if(thisPart > thatPart)
                return 1;
        }
        return 0;
    }

    @Override public boolean equals(Object that) {
        if(this == that)
            return true;
        if(that == null)
            return false;
        if(this.getClass() != that.getClass())
            return false;
        return this.compareTo((Version) that) == 0;
    }

}

Version a = new Version("1.1");
Version b = new Version("1.1.1");
a.compareTo(b) // return -1 (a<b)
a.equals(b)    // return false

Version a = new Version("2.0");
Version b = new Version("1.9.9");
a.compareTo(b) // return 1 (a>b)
a.equals(b)    // return false

Version a = new Version("1.0");
Version b = new Version("1");
a.compareTo(b) // return 0 (a=b)
a.equals(b)    // return true

Version a = new Version("1");
Version b = null;
a.compareTo(b) // return 1 (a>b)
a.equals(b)    // return false

List<Version> versions = new ArrayList<Version>();
versions.add(new Version("2"));
versions.add(new Version("1.0.5"));
versions.add(new Version("1.01.0"));
versions.add(new Version("1.00.1"));
Collections.min(versions).get() // return min version
Collections.max(versions).get() // return max version

// WARNING
Version a = new Version("2.06");
Version b = new Version("2.060");
a.equals(b)    // return false

编辑:

@daiscog:谢谢你的评论,这段代码是为Android平台开发的,由谷歌推荐,方法“匹配”检查整个字符串,不像Java使用监管模式。(Android文档- JAVA文档)

其他回答

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

public int compare(String v1, String v2) {
        v1 = v1.replaceAll("\\s", "");
        v2 = v2.replaceAll("\\s", "");
        String[] a1 = v1.split("\\.");
        String[] a2 = v2.split("\\.");
        List<String> l1 = Arrays.asList(a1);
        List<String> l2 = Arrays.asList(a2);


        int i=0;
        while(true){
            Double d1 = null;
            Double d2 = null;

            try{
                d1 = Double.parseDouble(l1.get(i));
            }catch(IndexOutOfBoundsException e){
            }

            try{
                d2 = Double.parseDouble(l2.get(i));
            }catch(IndexOutOfBoundsException e){
            }

            if (d1 != null && d2 != null) {
                if (d1.doubleValue() > d2.doubleValue()) {
                    return 1;
                } else if (d1.doubleValue() < d2.doubleValue()) {
                    return -1;
                }
            } else if (d2 == null && d1 != null) {
                if (d1.doubleValue() > 0) {
                    return 1;
                }
            } else if (d1 == null && d2 != null) {
                if (d2.doubleValue() > 0) {
                    return -1;
                }
            } else {
                break;
            }
            i++;
        }
        return 0;
    }

想知道为什么每个人都假设版本只由整数组成-在我的情况下,它不是。

为什么要重新发明轮子(假设版本遵循Semver标准)

首先通过Maven安装https://github.com/vdurmont/semver4j

然后使用这个库

Semver sem = new Semver("1.2.3");
sem.isGreaterThan("1.2.2"); // true

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'

这篇旧文章的另一个解决方案(对那些可能有帮助的人来说):

public class Version implements Comparable<Version> {

    private String version;

    public final String get() {
        return this.version;
    }

    public Version(String version) {
        if(version == null)
            throw new IllegalArgumentException("Version can not be null");
        if(!version.matches("[0-9]+(\\.[0-9]+)*"))
            throw new IllegalArgumentException("Invalid version format");
        this.version = version;
    }

    @Override public int compareTo(Version that) {
        if(that == null)
            return 1;
        String[] thisParts = this.get().split("\\.");
        String[] thatParts = that.get().split("\\.");
        int length = Math.max(thisParts.length, thatParts.length);
        for(int i = 0; i < length; i++) {
            int thisPart = i < thisParts.length ?
                Integer.parseInt(thisParts[i]) : 0;
            int thatPart = i < thatParts.length ?
                Integer.parseInt(thatParts[i]) : 0;
            if(thisPart < thatPart)
                return -1;
            if(thisPart > thatPart)
                return 1;
        }
        return 0;
    }

    @Override public boolean equals(Object that) {
        if(this == that)
            return true;
        if(that == null)
            return false;
        if(this.getClass() != that.getClass())
            return false;
        return this.compareTo((Version) that) == 0;
    }

}

Version a = new Version("1.1");
Version b = new Version("1.1.1");
a.compareTo(b) // return -1 (a<b)
a.equals(b)    // return false

Version a = new Version("2.0");
Version b = new Version("1.9.9");
a.compareTo(b) // return 1 (a>b)
a.equals(b)    // return false

Version a = new Version("1.0");
Version b = new Version("1");
a.compareTo(b) // return 0 (a=b)
a.equals(b)    // return true

Version a = new Version("1");
Version b = null;
a.compareTo(b) // return 1 (a>b)
a.equals(b)    // return false

List<Version> versions = new ArrayList<Version>();
versions.add(new Version("2"));
versions.add(new Version("1.0.5"));
versions.add(new Version("1.01.0"));
versions.add(new Version("1.00.1"));
Collections.min(versions).get() // return min version
Collections.max(versions).get() // return max version

// WARNING
Version a = new Version("2.06");
Version b = new Version("2.060");
a.equals(b)    // return false

编辑:

@daiscog:谢谢你的评论,这段代码是为Android平台开发的,由谷歌推荐,方法“匹配”检查整个字符串,不像Java使用监管模式。(Android文档- JAVA文档)