是否有比较版本号的标准习惯用法?我不能只使用直接的 String CompareTo,因为我还不知道最大的点释放数量是多少。我需要比较版本并确保以下内容成立:
1.0 < 1.1
1.0.1 < 1.1
1.9 < 1.10
这篇旧文章的另一个解决方案(对于那些可能有帮助的人):
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
Edit:
@daiscog:谢谢您的评论,这段代码是为Android平台开发的,并且按照Google的推荐,“匹配”方法检查整个字符串,这与Java使用监管模式不同。 (安卓文档 http://developer.android.com/reference/java/lang/String.html#matches%28java.lang.String%29 - JAVA文档 http://docs.oracle.com/javase/1.4.2/docs/api/java/lang/String.html#matches%28java.lang.String%29)
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