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

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

其他回答

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)

由于本页上没有答案能很好地处理混合文本,我做了自己的版本:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

class Main {
    static double parseVersion(String v) {
        if (v.isEmpty()) {
            return 0;
        }
        Pattern p = Pattern.compile("^(\\D*)(\\d*)(\\D*)$");
        Matcher m = p.matcher(v);
        m.find();
        if (m.group(2).isEmpty()) {
            // v1.0.0.[preview]
            return -1;
        }
        double i = Integer.parseInt(m.group(2));
        if (!m.group(3).isEmpty()) {
            // v1.0.[0b]
            i -= 0.1;
        }
        return i;
    }

    public static int versionCompare(String str1, String str2) {
        String[] v1 = str1.split("\\.");
        String[] v2 = str2.split("\\.");
        int i = 0;
        for (; i < v1.length && i < v2.length; i++) {
            double iv1 = parseVersion(v1[i]);
            double iv2 = parseVersion(v2[i]);

            if (iv1 != iv2) {
                return iv1 - iv2 < 0 ? -1 : 1;
            }
        }
        if (i < v1.length) {
            // "1.0.1", "1.0"
            double iv1 = parseVersion(v1[i]);
            return iv1 < 0 ? -1 : (int) Math.ceil(iv1);
        }
        if (i < v2.length) {
            double iv2 = parseVersion(v2[i]);
            return -iv2 < 0 ? -1 : (int) Math.ceil(iv2);
        }
        return 0;
    }


    public static void main(String[] args) {
        System.out.println("versionCompare(v1.0.0, 1.0.0)");
        System.out.println(versionCompare("v1.0.0", "1.0.0")); // 0

        System.out.println("versionCompare(v1.0.0b, 1.0.0)");
        System.out.println(versionCompare("v1.0.0b", "1.0.0")); // -1

        System.out.println("versionCompare(v1.0.0.preview, 1.0.0)");
        System.out.println(versionCompare("v1.0.0.preview", "1.0.0")); // -1

        System.out.println("versionCompare(v1.0, 1.0.0)");
        System.out.println(versionCompare("v1.0", "1.0.0")); // 0

        System.out.println("versionCompare(ver1.0, 1.0.1)");
        System.out.println(versionCompare("ver1.0", "1.0.1")); // -1
    }
}

不过,在需要比较“alpha”和“beta”的情况下,它仍然不够。

我自己写了一个小函数。更简单地使用列表

 public static boolean checkVersionUpdate(String olderVerison, String newVersion) {
        if (olderVerison.length() == 0 || newVersion.length() == 0) {
            return false;
        }
        List<String> newVerList = Arrays.asList(newVersion.split("\\."));
        List<String> oldVerList = Arrays.asList(olderVerison.split("\\."));

        int diff = newVerList.size() - oldVerList.size();
        List<String> newList = new ArrayList<>();
        if (diff > 0) {
            newList.addAll(oldVerList);
            for (int i = 0; i < diff; i++) {
                newList.add("0");
            }
            return examineArray(newList, newVerList, diff);
        } else if (diff < 0) {
            newList.addAll(newVerList);
            for (int i = 0; i < -diff; i++) {
                newList.add("0");
            }
            return examineArray(oldVerList, newList, diff);
        } else {
            return examineArray(oldVerList, newVerList, diff);
        }

    }

    public static boolean examineArray(List<String> oldList, List<String> newList, int diff) {
        boolean newVersionGreater = false;
        for (int i = 0; i < oldList.size(); i++) {
            if (Integer.parseInt(newList.get(i)) > Integer.parseInt(oldList.get(i))) {
                newVersionGreater = true;
                break;
            } else if (Integer.parseInt(newList.get(i)) < Integer.parseInt(oldList.get(i))) {
                newVersionGreater = false;
                break;
            } else {
                newVersionGreater = diff > 0;
            }
        }

        return newVersionGreater;
    }

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
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;
    }