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你如何比较Java中的两个版本字符串？

作者：罗文彬2502852027 | 2023-09-04 10:03
如何解决《你如何比较Java中的两个版本字符串？》经验，为你挑选了8个好方法。

是否有比较版本号的标准习惯用法？我不能只使用直接的字符串比较,因为我还不知道最大点数发布的数量.我需要比较版本,并具有以下成立:

1.0 < 1.1
1.0.1 < 1.1
1.9 < 1.10

alex.. 168

这个旧帖子的另一个解决方案(对于那些可能有帮助的人):

public class Version implements Comparable {

    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 (ab)
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 versions = new ArrayList();
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:感谢您的评论,这段代码是针对Android平台开发的,并且根据Google的推荐,"匹配"方法检查整个字符串,而不像使用监管模式的Java.(Android文档 - JAVA文档)



1> alex..：

这个旧帖子的另一个解决方案(对于那些可能有帮助的人):

public class Version implements Comparable {

    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 (ab)
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 versions = new ArrayList();
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:感谢您的评论,这段代码是针对Android平台开发的,并且根据Google的推荐,"匹配"方法检查整个字符串,而不像使用监管模式的Java.(Android文档 - JAVA文档)

高明!+1这应该是接受的答案

2> Alex Dean..：

使用Maven非常简单:

import org.apache.maven.artifact.versioning.DefaultArtifactVersion;

DefaultArtifactVersion minVersion = new DefaultArtifactVersion("1.0.1");
DefaultArtifactVersion maxVersion = new DefaultArtifactVersion("1.10");

DefaultArtifactVersion version = new DefaultArtifactVersion("1.11");

if (version.compareTo(minVersion) < 0 || version.compareTo(maxVersion) > 0) {
    System.out.println("Sorry, your version is unsupported");
}

您可以从此页面获取Maven Artifact的正确依赖关系字符串:


org.apache.maven
maven-artifact
3.0.3

请注意,`Comparable.compareTo`被记录为返回"负整数,零或正整数",因此避免检查-1和+1是一个好习惯.
完美,不要重新发明轮子!
只有一个担心是:在其中使用这种依赖关系,只有一个原因 - 拥有一个类 - DefaultArtifactVersion
@ses存储很便宜 - 当然比编写,测试和维护原始代码便宜
我已经创建了一个关于如何做到这一点的测试的要点:https://gist.github.com/2627608
这是完全错误的.`DefaultArtifactVersion`中的`compareTo`经常返回0,-1,1之外的值,因为它从另一个版本号中减去一个版本号.在这种情况下,这段代码将失败.看到不正确的答案赢得如此多的赞成票令人失望.

3> gizmo..：

使用点作为分隔符对字符串进行标记,然后从左侧开始并排比较整数平移.

并且不要忘记你可能并不总是只有数字.一些应用程序将包含内部版本号,并可能包含1.0.1b for beta/etc等内容.
你怎么做到这一点？

4> Peter Lawrey..：

您需要规范化版本字符串,以便对它们进行比较.就像是

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'

注意到标准化是你在那里隐含的最大宽度.
此代码导致'4.1.0'>'4.1'

5> 小智..：

最好重用现有代码,使用Maven的ComparableVersion类

好处:

Apache许可证,版本2.0,

测试,

在spring-security-core,jboss等多个项目中使用(复制)

多种功能

它已经是java.lang.Comparable了

只复制粘贴一个类,没有第三方依赖

不要包含对maven-artifact的依赖,因为这会引发各种传递依赖

这与问题相关,因为它是关于比较版本和maven版本比较的标准方法是非常标准的.
这是最好的答案.我无法相信有多少人(包括被接受的人)在没有测试的情况下尝试一些hacky字符串拆分.使用此类的代码示例:`assertTrue(new ComparableVersion("1.1-BETA").compareTo(new ComparableVersion("1.1-RC"))<0)`

6> Markus Jarde..：
// VersionComparator.java
import java.util.Comparator;

public class VersionComparator implements Comparator {

    public boolean equals(Object o1, Object o2) {
        return compare(o1, o2) == 0;
    }

    public int compare(Object o1, Object o2) {
        String version1 = (String) o1;
        String version2 = (String) o2;

        VersionTokenizer tokenizer1 = new VersionTokenizer(version1);
        VersionTokenizer tokenizer2 = new VersionTokenizer(version2);

        int number1 = 0, number2 = 0;
        String suffix1 = "", suffix2 = "";

        while (tokenizer1.MoveNext()) {
            if (!tokenizer2.MoveNext()) {
                do {
                    number1 = tokenizer1.getNumber();
                    suffix1 = tokenizer1.getSuffix();
                    if (number1 != 0 || suffix1.length() != 0) {
                        // Version one is longer than number two, and non-zero
                        return 1;
                    }
                }
                while (tokenizer1.MoveNext());

                // Version one is longer than version two, but zero
                return 0;
            }

            number1 = tokenizer1.getNumber();
            suffix1 = tokenizer1.getSuffix();
            number2 = tokenizer2.getNumber();
            suffix2 = tokenizer2.getSuffix();

            if (number1 < number2) {
                // Number one is less than number two
                return -1;
            }
            if (number1 > number2) {
                // Number one is greater than number two
                return 1;
            }

            boolean empty1 = suffix1.length() == 0;
            boolean empty2 = suffix2.length() == 0;

            if (empty1 && empty2) continue; // No suffixes
            if (empty1) return 1; // First suffix is empty (1.2 > 1.2b)
            if (empty2) return -1; // Second suffix is empty (1.2a < 1.2)

            // Lexical comparison of suffixes
            int result = suffix1.compareTo(suffix2);
            if (result != 0) return result;

        }
        if (tokenizer2.MoveNext()) {
            do {
                number2 = tokenizer2.getNumber();
                suffix2 = tokenizer2.getSuffix();
                if (number2 != 0 || suffix2.length() != 0) {
                    // Version one is longer than version two, and non-zero
                    return -1;
                }
            }
            while (tokenizer2.MoveNext());

            // Version two is longer than version one, but zero
            return 0;
        }
        return 0;
    }
}

// VersionTokenizer.java
public class VersionTokenizer {
    private final String _versionString;
    private final int _length;

    private int _position;
    private int _number;
    private String _suffix;
    private boolean _hasValue;

    public int getNumber() {
        return _number;
    }

    public String getSuffix() {
        return _suffix;
    }

    public boolean hasValue() {
        return _hasValue;
    }

    public VersionTokenizer(String versionString) {
        if (versionString == null)
            throw new IllegalArgumentException("versionString is null");

        _versionString = versionString;
        _length = versionString.length();
    }

    public boolean MoveNext() {
        _number = 0;
        _suffix = "";
        _hasValue = false;

        // No more characters
        if (_position >= _length)
            return false;

        _hasValue = true;

        while (_position < _length) {
            char c = _versionString.charAt(_position);
            if (c < '0' || c > '9') break;
            _number = _number * 10 + (c - '0');
            _position++;
        }

        int suffixStart = _position;

        while (_position < _length) {
            char c = _versionString.charAt(_position);
            if (c == '.') break;
            _position++;
        }

        _suffix = _versionString.substring(suffixStart, _position);

        if (_position < _length) _position++;

        return true;
    }
}

例:

public class Main
{
    private static VersionComparator cmp;

    public static void main (String[] args)
    {
        cmp = new VersionComparator();
        Test(new String[]{"1.1.2", "1.2", "1.2.0", "1.2.1", "1.12"});
        Test(new String[]{"1.3", "1.3a", "1.3b", "1.3-SNAPSHOT"});
    }

    private static void Test(String[] versions) {
        for (int i = 0; i < versions.length; i++) {
            for (int j = i; j < versions.length; j++) {
                Test(versions[i], versions[j]);
            }
        }
    }

    private static void Test(String v1, String v2) {
        int result = cmp.compare(v1, v2);
        String op = "==";
        if (result < 0) op = "<";
        if (result > 0) op = ">";
        System.out.printf("%s %s %s\n", v1, op, v2);
    }
}

输出:

1.1.2 == 1.1.2                --->  same length and value
1.1.2 < 1.2                   --->  first number (1) less than second number (2) => -1
1.1.2 < 1.2.0                 --->  first number (1) less than second number (2) => -1
1.1.2 < 1.2.1                 --->  first number (1) less than second number (2) => -1
1.1.2 < 1.12                  --->  first number (1) less than second number (12) => -1
1.2 == 1.2                    --->  same length and value
1.2 == 1.2.0                  --->  first shorter than second, but zero
1.2 < 1.2.1                   --->  first shorter than second, and non-zero
1.2 < 1.12                    --->  first number (2) less than second number (12) => -1
1.2.0 == 1.2.0                --->  same length and value
1.2.0 < 1.2.1                 --->  first number (0) less than second number (1) => -1
1.2.0 < 1.12                  --->  first number (2) less than second number (12) => -1
1.2.1 == 1.2.1                --->  same length and value
1.2.1 < 1.12                  --->  first number (2) less than second number (12) => -1
1.12 == 1.12                  --->  same length and value

1.3 == 1.3                    --->  same length and value
1.3 > 1.3a                    --->  first suffix ('') is empty, but not second ('a') => 1
1.3 > 1.3b                    --->  first suffix ('') is empty, but not second ('b') => 1
1.3 > 1.3-SNAPSHOT            --->  first suffix ('') is empty, but not second ('-SNAPSHOT') => 1
1.3a == 1.3a                  --->  same length and value
1.3a < 1.3b                   --->  first suffix ('a') compared to second suffix ('b') => -1
1.3a < 1.3-SNAPSHOT           --->  first suffix ('a') compared to second suffix ('-SNAPSHOT') => -1
1.3b == 1.3b                  --->  same length and value
1.3b < 1.3-SNAPSHOT           --->  first suffix ('b') compared to second suffix ('-SNAPSHOT') => -1
1.3-SNAPSHOT == 1.3-SNAPSHOT  --->  same length and value



7> Alex..：

想知道为什么每个人都认为这些版本只是由整数组成 - 在我的例子中并非如此.

为什么重新发明轮子(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



8> Algorithmati..：
public static int compareVersions(String version1, String version2){

    String[] levels1 = version1.split("\\.");
    String[] levels2 = version2.split("\\.");

    int length = Math.max(levels1.length, levels2.length);
    for (int i = 0; i < length; i++){
        Integer v1 = i < levels1.length ? Integer.parseInt(levels1[i]) : 0;
        Integer v2 = i < levels2.length ? Integer.parseInt(levels2[i]) : 0;
        int compare = v1.compareTo(v2);
        if (compare != 0){
            return compare;
        }
    }

    return 0;
}

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