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
如果你的项目中已经有Jackson,你可以使用com.fasterxml.jackson.core.Version:
import com.fasterxml.jackson.core.Version;
import org.junit.Test;
import static org.junit.Assert.assertTrue;
public class VersionTest {
@Test
public void shouldCompareVersion() {
Version version1 = new Version(1, 11, 1, null, null, null);
Version version2 = new Version(1, 12, 1, null, null, null);
assertTrue(version1.compareTo(version2) < 0);
}
}
其他回答
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!
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
我写了一个名为MgntUtils的开源库,它有一个用于字符串版本的实用程序。它正确地比较它们,适用于版本范围等等。下面是这个库javadoc参见方法TextUtils.comapreVersions(…)它已经被大量使用并经过了良好的测试。下面这篇文章描述了这个库以及如何获取它。它可以作为Maven工件和在github上获得(包括源代码和JavaDoc)
public static int compareVersions(String version1, String version2){
String[] levels1 = version1.split("\\.");
String[] levels2 = version2.split("\\.");
int length = Math.max(levels1.length, levels2.length);
for (int i = 0; i < length; i++){
Integer v1 = i < levels1.length ? Integer.parseInt(levels1[i]) : 0;
Integer v2 = i < levels2.length ? Integer.parseInt(levels2[i]) : 0;
int compare = v1.compareTo(v2);
if (compare != 0){
return compare;
}
}
return 0;
}
我现在就做了,然后问自己,这对吗?因为我从来没有找到过比我的更干净的解决方案
你只需要像下面这样拆分字符串版本("1.0.0"):
userVersion.split("\\.")
那么你将得到:{"1","0","0"}
现在,用我做过的方法
isUpdateAvailable(userVersion.split("\\."), latestVersionSplit.split("\\."));
方法:
/**
* Compare two versions
*
* @param userVersionSplit - User string array with major, minor and patch version from user (exemple: {"5", "2", "70"})
* @param latestVersionSplit - Latest string array with major, minor and patch version from api (example: {"5", "2", "71"})
* @return true if user version is smaller than latest version
*/
public static boolean isUpdateAvailable(String[] userVersionSplit, String[] latestVersionSplit) {
try {
int majorUserVersion = Integer.parseInt(userVersionSplit[0]);
int minorUserVersion = Integer.parseInt(userVersionSplit[1]);
int patchUserVersion = Integer.parseInt(userVersionSplit[2]);
int majorLatestVersion = Integer.parseInt(latestVersionSplit[0]);
int minorLatestVersion = Integer.parseInt(latestVersionSplit[1]);
int patchLatestVersion = Integer.parseInt(latestVersionSplit[2]);
if (majorUserVersion <= majorLatestVersion) {
if (majorUserVersion < majorLatestVersion) {
return true;
} else {
if (minorUserVersion <= minorLatestVersion) {
if (minorUserVersion < minorLatestVersion) {
return true;
} else {
return patchUserVersion < patchLatestVersion;
}
}
}
}
} catch (Exception ignored) {
// Will be throw only if the versions pattern is different from "x.x.x" format
// Will return false at the end
}
return false;
}
等待任何反馈:)