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
我写了一个小的Java/Android库来比较版本号:https://github.com/G00fY2/version-compare
它的基本功能是:
public int compareVersions(String versionA, String versionB) {
String[] versionTokensA = versionA.split("\\.");
String[] versionTokensB = versionB.split("\\.");
List<Integer> versionNumbersA = new ArrayList<>();
List<Integer> versionNumbersB = new ArrayList<>();
for (String versionToken : versionTokensA) {
versionNumbersA.add(Integer.parseInt(versionToken));
}
for (String versionToken : versionTokensB) {
versionNumbersB.add(Integer.parseInt(versionToken));
}
final int versionASize = versionNumbersA.size();
final int versionBSize = versionNumbersB.size();
int maxSize = Math.max(versionASize, versionBSize);
for (int i = 0; i < maxSize; i++) {
if ((i < versionASize ? versionNumbersA.get(i) : 0) > (i < versionBSize ? versionNumbersB.get(i) : 0)) {
return 1;
} else if ((i < versionASize ? versionNumbersA.get(i) : 0) < (i < versionBSize ? versionNumbersB.get(i) : 0)) {
return -1;
}
}
return 0;
}
这个代码片段不提供任何错误检查或处理。除此之外,我的库还支持像“1.2-rc”>“1.2-beta”这样的后缀。
其他回答
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;
}
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;
}
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。
我喜欢@Peter Lawrey的想法,我把它扩展到更远的范围:
/**
* Normalize string array,
* Appends zeros if string from the array
* has length smaller than the maxLen.
**/
private String normalize(String[] split, int maxLen){
StringBuilder sb = new StringBuilder("");
for(String s : split) {
for(int i = 0; i<maxLen-s.length(); i++) sb.append('0');
sb.append(s);
}
return sb.toString();
}
/**
* Removes trailing zeros of the form '.00.0...00'
* (and does not remove zeros from, say, '4.1.100')
**/
public String removeTrailingZeros(String s){
int i = s.length()-1;
int k = s.length()-1;
while(i >= 0 && (s.charAt(i) == '.' || s.charAt(i) == '0')){
if(s.charAt(i) == '.') k = i-1;
i--;
}
return s.substring(0,k+1);
}
/**
* Compares two versions(works for alphabets too),
* Returns 1 if v1 > v2, returns 0 if v1 == v2,
* and returns -1 if v1 < v2.
**/
public int compareVersion(String v1, String v2) {
// Uncomment below two lines if for you, say, 4.1.0 is equal to 4.1
// v1 = removeTrailingZeros(v1);
// v2 = removeTrailingZeros(v2);
String[] splitv1 = v1.split("\\.");
String[] splitv2 = v2.split("\\.");
int maxLen = 0;
for(String str : splitv1) maxLen = Math.max(maxLen, str.length());
for(String str : splitv2) maxLen = Math.max(maxLen, str.length());
int cmp = normalize(splitv1, maxLen).compareTo(normalize(splitv2, maxLen));
return cmp > 0 ? 1 : (cmp < 0 ? -1 : 0);
}
希望它能帮助到别人。它通过了interviewbit和leetcode中的所有测试用例(需要取消compareVersion函数中的两行注释)。
很容易测试!
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>