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
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;
}