在Java中比较版本字符串的有效方法 [英] Efficient way to compare version strings in Java

问题描述

可能重复：

Possible Duplicate:
How do you compare two version Strings in Java?

我有两个字符串，其中包含如下所示的版本信息：

I've 2 strings which contains version information as shown below:

str1 = "1.2"
str2 = "1.1.2"

现在，任何人都能告诉我有效的方式比较这些版本在Java&返回0，如果它们相等，-1，如果str1 < str2& 1如果str1> str2。

Now, can any one tell me the efficient way to compare these versions inside strings in Java & return 0 , if they're equal, -1, if str1 < str2 & 1 if str1>str2.

推荐答案

/**
 * Compares two version strings. 
 * 
 * Use this instead of String.compareTo() for a non-lexicographical 
 * comparison that works for version strings. e.g. "1.10".compareTo("1.6").
 * 
 * @note It does not work if "1.10" is supposed to be equal to "1.10.0".
 * 
 * @param str1 a string of ordinal numbers separated by decimal points. 
 * @param str2 a string of ordinal numbers separated by decimal points.
 * @return The result is a negative integer if str1 is _numerically_ less than str2. 
 *         The result is a positive integer if str1 is _numerically_ greater than str2. 
 *         The result is zero if the strings are _numerically_ equal.
 */
public static int versionCompare(String str1, String str2) {
    String[] vals1 = str1.split("\\.");
    String[] vals2 = str2.split("\\.");
    int i = 0;
    // set index to first non-equal ordinal or length of shortest version string
    while (i < vals1.length && i < vals2.length && vals1[i].equals(vals2[i])) {
      i++;
    }
    // compare first non-equal ordinal number
    if (i < vals1.length && i < vals2.length) {
        int diff = Integer.valueOf(vals1[i]).compareTo(Integer.valueOf(vals2[i]));
        return Integer.signum(diff);
    }
    // the strings are equal or one string is a substring of the other
    // e.g. "1.2.3" = "1.2.3" or "1.2.3" < "1.2.3.4"
    return Integer.signum(vals1.length - vals2.length);
}

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