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

我现在就做了,然后问自己,这对吗?因为我从来没有找到过比我的更干净的解决方案

你只需要像下面这样拆分字符串版本("1.0.0"):

userVersion.split("\\.")

那么你将得到:{"1","0","0"}

现在,用我做过的方法

isUpdateAvailable(userVersion.split("\\."), latestVersionSplit.split("\\."));

方法:

/**
 * Compare two versions
 *
 * @param userVersionSplit   - User string array with major, minor and patch version from user (exemple: {"5", "2", "70"})
 * @param latestVersionSplit - Latest string array with major, minor and patch version from api (example: {"5", "2", "71"})
 * @return true if user version is smaller than latest version
 */
public static boolean isUpdateAvailable(String[] userVersionSplit, String[] latestVersionSplit) {

    try {
        int majorUserVersion = Integer.parseInt(userVersionSplit[0]);
        int minorUserVersion = Integer.parseInt(userVersionSplit[1]);
        int patchUserVersion = Integer.parseInt(userVersionSplit[2]);

        int majorLatestVersion = Integer.parseInt(latestVersionSplit[0]);
        int minorLatestVersion = Integer.parseInt(latestVersionSplit[1]);
        int patchLatestVersion = Integer.parseInt(latestVersionSplit[2]);

        if (majorUserVersion <= majorLatestVersion) {
            if (majorUserVersion < majorLatestVersion) {
                return true;
            } else {
                if (minorUserVersion <= minorLatestVersion) {
                    if (minorUserVersion < minorLatestVersion) {
                        return true;
                    } else {
                        return patchUserVersion < patchLatestVersion;
                    }
                }
            }
        }
    } catch (Exception ignored) {
        // Will be throw only if the versions pattern is different from "x.x.x" format
        // Will return false at the end
    }

    return false;
}

等待任何反馈:)

其他回答

用点作为分隔符对字符串进行标记,然后从左边开始并排比较整数转换。

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'

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。

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;
}

对于我的项目,我使用我的公共版本库https://github.com/raydac/commons-version 它包含两个辅助类-用于解析版本(解析后的版本可以与另一个版本对象进行比较,因为它是可比的)和VersionValidator,它允许检查一些表达式的版本,如!=ide-1.1.1,>idea-1.3.4-SNAPSHOT;<1.2.3