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
使用Maven真的很简单:
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");
}
您可以从这个页面获得Maven Artifact的正确依赖项字符串:
<dependency>
<groupId>org.apache.maven</groupId>
<artifactId>maven-artifact</artifactId>
<version>3.0.3</version>
</dependency>
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。
使用Maven真的很简单:
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");
}
您可以从这个页面获得Maven Artifact的正确依赖项字符串:
<dependency>
<groupId>org.apache.maven</groupId>
<artifactId>maven-artifact</artifactId>
<version>3.0.3</version>
</dependency>
使用Java 8 Stream替换组件中的前导零。这段代码通过了interviewbit.com上的所有测试
public int compareVersion(String A, String B) {
List<String> strList1 = Arrays.stream(A.split("\\."))
.map(s -> s.replaceAll("^0+(?!$)", ""))
.collect(Collectors.toList());
List<String> strList2 = Arrays.stream(B.split("\\."))
.map(s -> s.replaceAll("^0+(?!$)", ""))
.collect(Collectors.toList());
int len1 = strList1.size();
int len2 = strList2.size();
int i = 0;
while(i < len1 && i < len2){
if (strList1.get(i).length() > strList2.get(i).length()) return 1;
if (strList1.get(i).length() < strList2.get(i).length()) return -1;
int result = new Long(strList1.get(i)).compareTo(new Long(strList2.get(i)));
if (result != 0) return result;
i++;
}
while (i < len1){
if (!strList1.get(i++).equals("0")) return 1;
}
while (i < len2){
if (!strList2.get(i++).equals("0")) return -1;
}
return 0;
}
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;
}
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;
}