Regular expression - parse version number string

Question

Hi,
I try to parse a string using regular expressions but id did not work correctly in all cases.
The string that i want to parse can have one of the following layout:
3.4.5.v20090305 or
3.4.5The first three parts have to be integer values, the last part can be a free string (which is optional).
I use the following code to parse the version number and extract the parts:
Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
Matcher m = versionPattern.matcher(versionString);
if (!m.find()) {
    throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// assert that we matched every part
// three groups (without qualifier) or four parts (with qualifier)
int groups = m.groupCount();
if (groups != 4) {
     throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// extract the parts
major = parseInt(m.group(1));
minor = parseInt(m.group(2));
micro = parseInt(m.group(3));
qualifier = m.group(4);The above regular expression works in all cases that i tested, except one.
The follwoing string is accepted as correct input: (but it shouldn't)
3.4.5a.v20090305And i get the result:
major = 3
minor = 4
micro = 5
qualifier = a.v20090305
Thanks for help or suggestions :)
Best Regards,
Michael

Answer 1

heissm wrote:
Hi,
I try to parse a string using regular expressions but id did not work correctly in all cases.
The string that i want to parse can have one of the following layout:
3.4.5.v20090305 or
3.4.5The first three parts have to be integer values, the last part can be a free string (which is optional).
I use the following code to parse the version number and extract the parts:
Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
Matcher m = versionPattern.matcher(versionString);
if (!m.find()) {
throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// assert that we matched every part
// three groups (without qualifier) or four parts (with qualifier)
int groups = m.groupCount();
if (groups != 4) {
throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// extract the parts
major = parseInt(m.group(1));
minor = parseInt(m.group(2));
micro = parseInt(m.group(3));
qualifier = m.group(4);The above regular expression works in all cases that i tested, except one.
The follwoing string is accepted as correct input: (but it shouldn't)
3.4.5a.v20090305And i get the result:
major = 3
minor = 4
micro = 5
qualifier = a.v20090305No, that can't be the output. Perhaps you have some old class files you're executing, because with the code you now posted, that can't be the result.
To verify this, execute this:
import java.util.regex.*;
public class Main {
    public static void main(String[] args) {
        String versionString = "3.4.5a.v20090305";
        Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
        Matcher m = versionPattern.matcher(versionString);
        if (!m.find()) {
            throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
        // assert that we matched every part
        // three groups (without qualifier) or four parts (with qualifier)
        int groups = m.groupCount();
        if (groups != 4) {
             throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
        // extract the parts
        System.out.println("1 -> "+m.group(1));
        System.out.println("2 -> "+m.group(2));
        System.out.println("3 -> "+m.group(3));
        System.out.println("4 -> "+m.group(4));
}and you'll see that the following is the result:
1 -> 3
2 -> 4
3 -> 5
4 -> nullSome remarks about your pattern:
"(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?"All the "{1}" can be omitted they don't add anything to your regex and are only cluttering it up. And you'll probably also want to "anchor" your regex using the start- and end-of-string-meta-characters, like this:
"^(\\d+)\\.(\\d+)\\.(\\d+)(?:\\.(\\w+))?$"