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
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;
}
其他回答
我创建了一个简单的实用程序,使用语义版本约定在Android平台上比较版本。所以它只适用于X.Y.Z (Major.Minor.Patch)格式的字符串,其中X、Y和Z是非负整数。你可以在我的GitHub上找到它。
方法version . compareversions (String v1, String v2)比较两个版本字符串。如果版本相等则返回0,如果版本v1在版本v2之前则返回1,如果版本v1在版本v2之后则返回-1,如果版本格式无效则返回-2。
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)
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;
}
基于https://stackoverflow.com/a/27891752/2642478
class Version(private val value: String) : Comparable<Version> {
private val splitted by lazy { value.split("-").first().split(".").map { it.toIntOrNull() ?: 0 } }
override fun compareTo(other: Version): Int {
for (i in 0 until maxOf(splitted.size, other.splitted.size)) {
val compare = splitted.getOrElse(i) { 0 }.compareTo(other.splitted.getOrElse(i) { 0 })
if (compare != 0)
return compare
}
return 0
}
}
你可以用like:
System.err.println(Version("1.0").compareTo( Version("1.0")))
System.err.println(Version("1.0") < Version("1.1"))
System.err.println(Version("1.10") > Version("1.9"))
System.err.println(Version("1.10.1") > Version("1.10"))
System.err.println(Version("0.0.1") < Version("1"))
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;
}