java UUID的java正则表达式
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java regex for UUID
提问by Aqura
I want to parse a String which has UUID in the below format
我想解析一个具有以下格式的 UUID 的字符串
"<urn:uuid:4324e9d5-8d1f-442c-96a4-6146640da7ce>"
I have tried it parsing in below way, which works, however I think it would be slow
我已经尝试以下面的方式解析它，这有效，但是我认为它会很慢
private static final String reg1 = ".*?";
private static final String reg2 = "([A-Z0-9]{8}-[A-Z0-9]{4}-[A-Z0-9]{4}-[A-Z0-9]{4}-[A-Z0-9]{12})";
private static final Pattern splitter = Pattern.compile(re1 + re2, Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
I am looking for a faster way and tried below, but it fails to match
我正在寻找一种更快的方法并在下面尝试过，但它不匹配
private static final Pattern URN_UUID_PATTERN = Pattern.compile("^< urn:uuid:([^&])+>");
I am new to regex. any help is appreciated.
我是正则表达式的新手。任何帮助表示赞赏。
\Aqura
\阿库拉
回答by dlamblin
Your example of a faster regex is using a <where the input is <so that's confusing.
您的更快正则表达式示例使用的<是输入位置，<因此令人困惑。
Regarding speed, first, your UUID is hexadecimal, so don't match with A-Zbut rather a-f. Second you give no indication that case is mixed, so don't use case insensitive and write the correct case in the range.
关于速度，首先，你的 UUID 是十六进制的，所以不要匹配A-Z而是a-f. 其次，您没有给出大小写混合的迹象，因此不要使用不区分大小写的方式并在范围内写入正确的大小写。
You don't explain if you need the part preceding the UUID. If not, don't include .*?, and you may as well write the literals for re1and re2together in your final Pattern. There's no indication you need DOTALL either.
您没有解释是否需要 UUID 前面的部分。如果不是，不包括.*?，你可能也写的文字re1和re2你在一起final Pattern。也没有迹象表明您需要 DOTALL。
private static final Pattern splitter =
Pattern.compile("([a-f0-9]{8}(-[a-f0-9]{4}){4}[a-f0-9]{8})");
Alternatively, if you are measuring your Regular Expression's performance to be too slow, you might try another approach, for example:
Is each uuid preceded by "uuid:" as in your example? If so you can
或者，如果您测量正则表达式的性能太慢，您可以尝试另一种方法，例如：
每个 uuid 前面是否与示例中的“uuid:”一样？如果是这样你可以
- find the first index of "uuid:" as i, then
- substring 0 to i+5 [assuming you needed it at all], and
- substring i+5 to i+41, if I counted that right (36 characters in length).
- 找到 "uuid:" 的第一个索引作为i，然后
- 子串 0 到i+5 [假设你需要它]，和
- 子字符串i+5 到i+41，如果我算对了（长度为 36 个字符）。
Along similar lines your faster regex could be:
沿着类似的路线，您更快的正则表达式可能是：
private static final Pattern URN_UUID_PATTERN =
Pattern.compile("^<urn:uuid:(.{36})>");
OTOH if all your input strings are going to start with those exact characters, no need to do step 1 in the previous suggestion, just input.substring(13, 49);
OTOH 如果您的所有输入字符串都以这些确切字符开头，则无需执行上一个建议中的第 1 步，只需 input.substring(13, 49);
回答by Alexander du Sautoy
If this format don't be changed. I think more fast way is use String.substring() method. Example:
如果这种格式不改变。我认为更快捷的方法是使用 String.substring() 方法。例子：
String val = "<urn:uuid:4324e9d5-8d1f-442c-96a4-6146640da7ce>";
String sUuid = val.substring(13, 49);
UUID uuid = UUID.fromString(sUuid);
Inside class String used char array for store data, in package java.lang.String:
在类 String 内部使用字符数组存储数据，在包 java.lang.String 中：
public final class String
implements java.io.Serializable, Comparable<String>, CharSequence {
...
113: /** The value is used for character storage. */
114: private final char value[];
...
}
Method 'String substring(int beginIndex, int endIndex)' make the copy of array elements, from start to end index, and create new String on basis new array. Copying of array it is a very fast operation.
方法 'String substring(int beginIndex, int endIndex)' 复制数组元素，从开始到结束索引，并在新数组的基础上创建新字符串。复制数组是一个非常快的操作。
